Which expressions can be used to find the volume of the prism?

Answer:

[tex]\sqrt{53}[/tex]

Step-by-step explanation:

Answer:

760

Step-by-step explanation:

Answer:

7, 17, 37, 47, 73, 79, 97

Step-by-step explanation:

they have 7's as one of their digits

Answer:

Step-by-step explanation:

Find the total surface area of a right cylinder with a radius of 3 inches and a height of 17 inches. Round to the nearest tenth.

Your

Answer:

Step-by-step explanation:

suface area of base=1/2×9×7.8=9×3.9=35.1 yd²

lateral surface area=3×1/2×9×10=135 yd²

total surface area=135+35.1=170.1 yd²

i assumed it is an equilateral triangular base.

as √(7.8²+4.5²)=9.0049≈9

Answer:

Step-by-step explanation:

using pythagoras theorem

a^2 + b^2 = c^2

B^2 + 8^2 = 15^2

B^2 + 64 + 225

B^2 = 225 - 64

B^2 = 161

B = [tex]\sqrt{161}[/tex]    OR  12.7

Step-by-step explanation:

The volume of a cube is the edge³, so to find the edge, you need to calculate the cubic root of it:

[tex] \sqrt[3]{ \frac{1}{8} } = \frac{1}{2 } [/tex]

Answer:

The missing values are [tex]A = \frac{1}{4}[/tex], [tex]B = 1[/tex] and [tex]C = 4[/tex], respective.

Step-by-step explanation:

We find the values regarding A, B and C by evaluating [tex]g(x)[/tex] at respective values of [tex]x[/tex]:

[tex]x = -2[/tex]

[tex]g(-2) = 4^{-\frac{2}{2} }[/tex]

[tex]g(-2) = 4^{-1}[/tex]

[tex]g(-2) = \frac{1}{4}[/tex]

[tex]A = \frac{1}{4}[/tex]

[tex]x = 0[/tex]

[tex]g(0) = 4^{\frac{0}{2} }[/tex]

[tex]g(0) = 1[/tex]

[tex]B = 1[/tex]

[tex]x = 2[/tex]

[tex]g(2) = 4^{\frac{2}{2} }[/tex]

[tex]g(2) = 4^{1}[/tex]

[tex]g(2) = 4[/tex]

[tex]C = 4[/tex]

Answer:       A=1/4,   B=1,   C=4

Step-by-step explanation:

Correct on edge 2022

Answer:

-12, 0

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

The triangles are similar because they have the same angles.

Make a proportion to solve for AC which we will substitute with variable x.

AM = MB = 4

AB = AM + MB = 8

BN = NC = 3

BC = BN + NC = 6

big triangle sides: 8, 6, x

small triangle sides: 4, 3, 6

now we can write proportion.

8/4 = x/6

x = 12

AC = 12

Answer:

1/2 or 50% or .5

Step-by-step explanation:

She hits at .500 that's pretty good. I'd take her on my fantasy team.

Answer: pretty sure it is 201 tiles

please mark me brainlies. That would be helpful

Answer:

a23, a31, a21, a11, a12, a33

Step-by-step explanation: I got it right on Edumentum.

Answer:

a23, a31, a21, a11, a12, a33

Step-by-step explanation:

Step-by-step explanation:

y = 5^x -9

5^x = y + 9

x = ^5 log (y+9)

f~'(x) = ^5 log (x+9)

Answer:

Angle HFG = 24 degrees

Step-by-step explanation:

From the diagram, we have the following;

Angle EFH and angle HFG are complementary (they add up to be 90 degrees)

angle EFH + angle HFG = 90

Angle HFG = 90-66

Angle HFG = 24 degrees

Answer:

[tex]P = (3,0)[/tex]

Step-by-step explanation:

Given

[tex]A = (-3,-2)[/tex] ---- [tex](x_1,y_1)[/tex]

[tex]B = (6,1)[/tex] --- [tex](x_2,y_2)[/tex]

[tex]m:n =2:1[/tex]

Required

The coordinates of P

This is calculated as:

[tex]P = (\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}][/tex]

So:

[tex]P = (\frac{2*6+1*-3}{2+1},\frac{2*1+1*-2}{2+1})[/tex]

[tex]P = (\frac{9}{3},\frac{0}{3})[/tex]

[tex]P = (3,0)[/tex]

Answer:

s = 71

Step-by-step explanation:

Given that GJ is a midsegment of the triangle then it is half the length of the third side, that is

GJ = [tex]\frac{1}{2}[/tex] HI , so

s - 65 = [tex]\frac{1}{2}[/tex] (s - 59) ← multiply both sides by 2 to clear the fraction

2s - 130 = s - 59 ( subtract s from both sides )

s - 130 = - 59 ( add 130 to both sides )

s = 71

Answer:

9 dollars.

Step-by-step explanation:

A dozen is 12.

12/4 = 3

$3 * 3 = $9

Step-by-step explanation:

3×4=12. 12 is a dozen. $3=4 muffins. $6=8 muffins. 9=12 muffins . so $9 for a dozen

Consider IJ and GL are the parallel sides of the trapezoid.

Given:

In a trapezoid GIJL, HK is the mid segment, IJ =7 cm and GL=15 cm.

To find:

The length of the mid segment HK.

Solution:

The length of the mid segment of a trapezoid is:

[tex]\text{Midsegment}=\dfrac{b_1+b_2}{2}[/tex]

where, [tex]b_1,b_2[/tex] are two parallel side or bases of the trapezoid.

HK is the mid segment, and the parallel sides are IJ =7 cm and GL=15 cm. So, the length of the mid segment is:

[tex]HK=\dfrac{IJ+GL}{2}[/tex]

[tex]HK=\dfrac{7+15}{2}[/tex]

[tex]HK=\dfrac{22}{2}[/tex]

[tex]HK=11[/tex]

Therefore, the length of HK is 11 units.

Answer:

Gradient is change in y over change in x.

gradient = y1-y2 ÷ x1-x2

=7-11 ÷ 3-5

=-4 ÷ -2

=2

Use the slope/gradient formula below:

[tex] \large \boxed{m = \frac{y_2 - y_1}{x_2 - x_1} }[/tex]

This formula is for when we want to find the gradient when we have two given points. (We also call it Rise Over Run or changes of y over the changes of x.)

We are given two points which are (3,7) and (5,11). Substitute these points in the equation. (the ordered pairs are in (x,y) and not (y,x))

[tex] \large{m = \frac{11 - 7}{5 - 3} } \\ \large{m = \frac{4}{2} \longrightarrow \boxed{m = 2}}[/tex]

Hence, the gradient of AB is 2.

Answer

Step-by-step explanation:

The last two answers is correct

Answer:

4

Step-by-step explanation:

The answer to T - H is negative

Answer:

Step-by-step explanation:

If EG is a diameter, then arc EFG is a semicircle and its measure is 180. Arc FG then is 180 - 124 = 56. Since angle FEG is an inscribed angle and the arc it cuts off is arc FG, then the measure of the inscribed angle is half the measure of the arc it cuts off...so angle FEG is 28 degrees. Keep that in mind; we'll need it in a sec.

If HE is tangent to the circle at E, then angle HEG is a 90 degree angle. Adding that to angle FEG will give you angle FEH. Angle FEH = 90 + 28 = 118

=============================================================

Explanation:

Ignore the sides that are 'a' units long.

We'll compare the sides that are 26 units and 27 units long. We see the 27 unit side is larger, which makes the 93 degree angle larger than the 3x-9 degree angle.

Symbolically, we would say

3x-9 < 93

At the same time, the 3x-9 is also larger than 0

So

3x-9 > 0 or 0 < 3x-9

Put together, we can form this compound inequality

0 < 3x-9 < 93

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

Let's isolate x like so

0 < 3x-9 < 93

0+9 < 3x-9+9 < 93+9 .... add 9 to all sides

9 < 3x < 102

9/3 < 3x/3 < 102/3 .... divide all sides by 3

3 < x < 34

x is any number between 3 and 34, but cannot equal either endpoint

For example, x = 10 is valid since 3 < 10 < 34 is true. But x = 50 is not a solution since 3 < 50 < 34 is false. We can't have x = 3 and we also can't have x = 34 either, but we can have anything in between those endpoints.

Answer:

b = 6

Step-by-step explanation:

solve the second equation for 'y':

-3x + 1/2y = -3

add 3x to each side to get:

1/2y = 3x - 3

multiply all terms by 2 to eliminate the denominator to get:

y = 6x - 6

now compare '6x-6' with '6x-b'

you want both expressions to be equal because then there are an infinite number of solutions:

b = 6

Answer:

840

Step-by-step explanation:

600×140% =840 hope this helps

Answer:

the radius is 3 m

Step-by-step explanation:

the radius of a circle is always one half of the diameter.

Here the diameter is 6 m, so the radius is 3 m.

Answer:

1 pair of parallel sides: B

2 pairs of parallel sides: A, D

3 pairs of parallel sides: C

Step-by-step explanation:

When two lines are parallel, they will never intersect.

Answer:

Step-by-step explanation:

circumference of circle=2πr

=2*π*9

=18π m

[tex]{ \bf{ \underbrace{Given :}}}[/tex]

Radius of the circle = 9 m.

[tex]{ \bf{ \underbrace{To\:find:}}}[/tex]

The circumference of the circle.

[tex]{ \bf{ \underbrace{Solution :}}}[/tex]

A. C = 18 π ✅

[tex]\sf\orange{The\:circumference \:of\:the\:circle\:is\:18\:π\:m.}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

We know that,

[tex]\sf\purple{Circumference\:of\:a\:circle \:=\:2Πr }[/tex]

[tex] = 2 \: \pi \times 9 \: m \\ = 18 \: \pi \: m[/tex]

Therefore, the circumference of the circle is 18 π m.

[tex]\large\mathfrak{{\pmb{\underline{\blue{Happy\:learning }}{\blue{!}}}}}[/tex]