Help!!!!!!!Solve for x. Round to the nearest tenth of a degree, if necessary.

Answer:

das

Step-by-step explanation:

adsasc

Answer:

A. x > -4

Step-by-step explanation:

< less than

>greater than

the unknown number x is greater than -4

{9, 8, 7, ...} the number is counting down. So the next number is 6.

is 6 greater than -4

or less than -4

therefore

x(which is 6) is greater than -4

x > -4

Answer:

c

Step-by-step explanation:

Answer:

a. y=x-2     b. -2

Step-by-step explanation:

a.) if the slope is 1, the x-int is always the negative of the y-int so y=x-2

b.) in y=mb+b (slope intercept form) b is the y-int so -2

Brainliest Plz??

Answer:

[tex]4.\ \sin E = \cos G[/tex]

Step-by-step explanation:

Given

[tex]\triangle EFG[/tex]

[tex]\angle F = 90^o[/tex] --- right angle

Required

Which of the options is true

In a triangle, we have:

[tex]\angle E + \angle F + \angle G = 180^o[/tex] --- angles in a triangle

Substitute [tex]\angle F = 90^o[/tex]

[tex]\angle E + 90^o + \angle G = 180^o[/tex]

Collect like terms

[tex]\angle E + \angle G = 180^o -90^o[/tex]

[tex]\angle E + \angle G =90^o[/tex]

This implies that E and G are complementary angles.

For complementary angles, E and G;

[tex]\sin E = \cos G[/tex] and [tex]\sin G = \cos E[/tex]

Hence, (4) is true

Answer: 80%

Step-by-step explanation:

Hence, if the numerator is decreased by 40% and the denominator is decreased by 25%, the original fraction is decreased by 80 percent.

Answer:

B is the answer

Step-by-step explanation:

Answer:

A. 15.31

I did the standard deviation calculator and this is what I got

I think the answer should be 12

this is because CH length is 6 x 5 is 30 so 60/5 is 12 for BH.

Answer:

[tex]Mean = 4[/tex]

Step-by-step explanation:

Given

[tex]f(x) =\left \{ {{\frac{c}{x^3} \ x\ge 2} \atop {0\ x<2}} \right.[/tex]

Required

The mean of x

Given that:

[tex]f(x) = \frac{c}{x^3}[/tex]     [tex]x \ge 2[/tex]

First, solve for c using;

[tex]\int\limits^a_b {f(x)} \, dx = 1[/tex]

Substitute [tex]f(x) = \frac{c}{x^3}[/tex] and [tex]x \ge 2[/tex]

[tex]\int\limits^{\infty}_2 {\frac{c}{x^3}} \, dx = 1[/tex]

Isolate c

[tex]c\int\limits^{\infty}_2 {\frac{1}{x^3}} \, dx = 1[/tex]

Rewrite as:

[tex]c\int\limits^{\infty}_2 {x^{-3}} \, dx = 1[/tex]

[tex]c[\frac{x^{-3+1}}{-3 +1}]|\limits^{\infty}_2 = 1[/tex]

[tex]c[\frac{x^{-2}}{-2}]|\limits^{\infty}_2 = 1[/tex]

[tex]-\frac{c}{2} [x^{-2}]|\limits^{\infty}_2 = 1[/tex]

Expand

[tex]-\frac{c}{2} [{\infty}^{-2} - {2}^{-2}]= 1[/tex]

[tex]-\frac{c}{2} [0 - \frac{1}{4}]= 1[/tex]

[tex]-\frac{c}{2} *- \frac{1}{4}= 1[/tex]

[tex]\frac{c}{8}= 1[/tex]

Solve for c

[tex]x = 8 * 1[/tex]

[tex]x = 8[/tex]

So, we have:

[tex]f(x) = \frac{c}{x^3}[/tex]     [tex]x \ge 2[/tex]

[tex]f(x) = \frac{8}{x^3}[/tex]     [tex]x \ge 2[/tex]

So, the mean is calculated as:

[tex]Mean = \int\limits^a_b {x * f(x)} \, dx[/tex]

This gives;

[tex]Mean = \int\limits^{\infty}_2 {x * \frac{8}{x^3}} \, dx[/tex]

[tex]Mean = \int\limits^{\infty}_2 {\frac{8}{x^2}} \, dx[/tex]

[tex]Mean = 8\int\limits^{\infty}_2 {\frac{1}{x^2}} \, dx[/tex]

Rewrite as:

[tex]Mean = 8\int\limits^{\infty}_2 {x^{-2} \, dx[/tex]

Integrate

[tex]Mean = 8 {\frac{x^{-2+1}}{-2+1}|\limits^{\infty}_2[/tex]

[tex]Mean = 8 {\frac{x^{-1}}{-1}}|\limits^{\infty}_2[/tex]

[tex]Mean = -8x^{-1}|\limits^{\infty}_2[/tex]

Expand

[tex]Mean = -8[(\infty)^{-1} - 2^{-1}][/tex]

[tex]Mean = -8[0 - \frac{1}{2}][/tex]

[tex]Mean = -8* - \frac{1}{2}[/tex]

[tex]Mean = 8* \frac{1}{2}[/tex]

[tex]Mean = 4[/tex]

Answer:

B.

Step-by-step explanation:

3/12 reduces to 1/4.

1/2 alone is greater than 1/4, so when 2/10 is added to 1/2, it is certainly greater than 3/12.

Answer: B.

Answer:

I believe it's 17 :)

Step-by-step explanation:

5×3= 15 +2 = 17

Answer:

Max (-7/2, 97/4)

Domain: all real numbers

Range: (negative infinity, 97/4)

Step-by-step explanation:

Im assuming this is a quadratic equation y = -x^2-7x+12

The max/min are the vertex (-b/2a)

Answer:

2630 g

Step-by-step explanation:

From the given information:

Given that:

mean (μ) = 3750 g

Standard deviation (σ) = 500

Suppose the hospital officials demand special treatment with a percentage of  lightest 3% (0.03) for newborn babies;

Then, the weight of birth that differentiates the babies that needed special treatment from those that do not can be computed as follows;

P(Z < z₁) = 0.03

Using the Excel Formula; =NORMSINV(0.03) = -1.88

z₁ = - 1.88

Using the test statistics z₁ formula:

[tex]z_1 = \dfrac{X-\mu}{\sigma}[/tex]

[tex]-1.88 = \dfrac{X-3570}{500}[/tex]

By cross multiply, we have:

-1.88 × 500 = X - 3570

-940 = X - 3570

-X = -3570 + 940

-X = -2630

X = 2630 g

Hence, 2630 g is the required weight of birth   that differentiates the babies that needed special treatment from those that do not

Answer:

3/2

Step-by-step explanation:

(y-y)/(x-x)=-1+7/2+2=6/4=3/2

Answer:

Slope = 1.5

Step-by-step explanation:

m = (y₂ -y₁) / (x₂-x₁)

m = (-1+7) / (2+2)

m = 1.5

Answer:

[tex]S_{15}= 50[/tex]

Step-by-step explanation:

Given

[tex]a_7 = \frac{14}{3}[/tex]

[tex]d = -\frac{4}{3}[/tex]

[tex]n = 15[/tex]

Required

The sum of n terms

First, we calculate the first term using:

[tex]a_n = a + (n - 1)d[/tex]

Let [tex]n = 7[/tex]

So, we have:

[tex]a_7 = a + (7 - 1)d[/tex]

[tex]a_7 = a + 6d[/tex]

Substitute [tex]a_7 = \frac{14}{3}[/tex] and [tex]d = -\frac{4}{3}[/tex]

[tex]\frac{14}{3} = a + 6*\frac{-4}{3}[/tex]

[tex]\frac{14}{3} = a -8[/tex]

Collect like terms

[tex]a =\frac{14}{3} +8[/tex]

Take LCM and solve

[tex]a =\frac{14+24}{3}[/tex]

[tex]a =\frac{38}{3}[/tex]

The sum of n terms is then calculated as:

[tex]S_n = \frac{n}{2}(2a + (n - 1)d)[/tex]

Where: [tex]n = 15[/tex]

So, we have:

[tex]S_n = \frac{15}{2}(2*\frac{38}{3} + (15 - 1)*\frac{-4}{3})[/tex]

[tex]S_n = \frac{15}{2}(2*\frac{38}{3} + 14 *\frac{-4}{3})[/tex]

[tex]S_n = \frac{15}{2}(2*\frac{38}{3} - 14 *\frac{4}{3})[/tex]

[tex]S_n = \frac{15}{2}(\frac{2*38}{3} - \frac{14 *4}{3})[/tex]

Take LCM

[tex]S_n = \frac{15}{2}(\frac{2*38-14 *4}{3})[/tex]

[tex]S_n = \frac{15}{2}(\frac{20}{3})[/tex]

Open bracket

[tex]S_n = \frac{15*20}{2*3}[/tex]

[tex]S_n = \frac{300}{6}[/tex]

[tex]S_n = 50[/tex]

Hence,

[tex]S_{15}= 50[/tex]

Answer:

6. 8 in.²

7. 47.6 m²

9. 20 cm²

Step-by-step explanation:

6. Area of triangle = ½*base*height

base = 4 in.

height = 4 in.

Area = ½*4*4

Area = 2*4

Area = 8 in.²

7. Area of triangle = ½*base*height

base = 13.6 m

height = 7 m

Area = ½*13.6*7

Area = 47.6 m²

9. Area of the shaded figure = area of triangle + area of rectangle

= ½*b*h + L*W

b = 2 cm

h = 4 cm

L = 8 cm

W = 2 cm

Area of the shaded figure = ½*2*4 + 8*2

= 4 + 16

= 20 cm²

Answer:

42.39 ft²

Step-by-step explanation:

area of sector= 3/8 × 3.14 × 6²

= 42.39 ft²

The area of the sector is 42.39 ft²

Given that, a sector of a circle has a diameter of 12 feet and an angle of 3pi/4 radians, we need to find the area of the sector,

Area of the sector = θ/360° / π·radius²

area of sector= 3/8 × 3.14 × 6²

= 42.39 ft²

Hence, the  area of the sector is 42.39 ft²

Learn more about the sector of the circle, click;

https://brainly.com/question/15591260

#SPJ2

Answer:

Step-by-step explanation:

Perpendicular bisector of the chord connecting the the points (-1,2) and (2,3) is also passing through the center of the circle.

We'll find the equation of the line and solve the system to find the center.

Midpoint of the chord:

Equation of the line through chord (-1,2) and (2,3):

Perpendicular bisector is:

And the given line is:

Solve the system:

Find y:

The center is (-1, 7)

Find the radius, the distance from center to one of points on circle

The equation of circle is:

Hope This Helps!!!

Hope This Helps!!!Have A GREAT DAY!!!

Answer:

Step-by-step explanation:

13 - 6 = 6 - 1

7 = 5  (FALSE)

-----------------------------

6 - 5 = 7 - 6

1 = 1   (TRUE)

------------------------------

8 - 4 = 11 - 8

4 = 3  (FALSE)

Answer:

80

Step-by-step explanation:

because 16 decided by 0.2 is 80

hope this helps!!!!

Answer:

r = 1,248 in

Step-by-step explanation:

v(c) = 12 in³

The surface area of a right cylinder is:

Area of the base and top  + lateral area

S(a)  =  2*π*r²  + 2*π*r*h   (1)

v(c) = 12 in³  =  π*r²*h        h is the height of the cylinder, then

h  =  12 / π*r²

By substitution,  in equation (1) we get  the  Surface area as a function of r

S(r)  =  2*π*r² +   2*π*r*  ( 12 / π*r²)

S(r)  =  2*π*r² + 24 /r

Tacking derivatives on both sides of the equation we get:

S´(r)  =  4*π*r -  24 /r²

S´(r)  = 0            4*π*r -  24 /r²  =  0       π*r - 6/r² = 0

 π*r³ - 6  = 0

r³   =  1,91    

r = 1,248 in    

How do we know that the value  r  =  1,248 makes  Surface area minimum??

We get the second derivative

S´´(r)  =  4*π  +  48/r³    S´´(r)  will be always positive therefore we have a minumum of S   at the value of   r = 1,248 in

Answer: feugfberuyvfeygcreuyug

Step-by-step hththrhfhr

[tex]\huge\text{Hey there!}[/tex]

[tex]\large\text{If x = -2, then substitute it into the given equation}[/tex].

[tex]\large\textsf{f(x) = 4(-2) - 2}[/tex]

[tex]\large\textsf{y = 4(-2) - 2}[/tex]

[tex]\large\textsf{4(-2) = \boxed{\bf -8}}[/tex]

[tex]\large\textsf{y = -8 - 2}[/tex]

[tex]\large\textsf{-8 - 2 = \boxed{\bf -10}}[/tex]

[tex]\boxed{{\large\text{y = \bf -10}}}[/tex]

[tex]\boxed{\boxed{\large\text{Answer: \textsf{\huge \bf y = -10}}}}\huge\checkmark[/tex]

[tex]\large\text{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

[tex]\quad\quad\quad\quad \tt{f(x) = 4x - 2}[/tex]

[tex]\quad\quad\quad\quad \tt{f( - 2) = 4( - 2)- 2}[/tex]

[tex]\quad\quad\quad\quad \tt{f( - 2) = ( - 8)- 2}[/tex]

[tex]\quad\quad\quad\quad \tt{f( - 2) = - 10}[/tex]

[tex]\quad\quad\quad\quad\boxed {\tt{\color{green}f( - 2) = - 10}}[/tex]

__________

#LetsStudy

Answer:

perimeter = 25.62 units

Step-by-step explanation:

Let the quadrilateral be ABCD

A = (1, 5), B = (6, 5), C= (1, 11) , D = (6, 11)

To find the perimeter we need to add the lengths of sides AB , BC, CD, and AD.

Lets find the length of all sides using distance formula;

[tex]distance = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]

[tex]AB = \sqrt{(1-6)^2+(5-5)^2} = \sqrt{25} = 5units\\\\BC= \sqrt{(6-1)^2+(5-11)^2} = \sqrt{25+36} =\sqrt{61} units\\\\CD=\sqrt{(1-6)^2 + (11-11)^2} = \sqrt{25} = 5units\\\\AD = \sqrt{(1-6)^2+(5-11)^2} = \sqrt{25+36} = \sqrt{61}units[/tex]

Perimeter = AB + BC + CD + AD

               [tex]=10 + 2\sqrt{61} units[/tex] = 25.62 units