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Pavlova-9 [17]
3 years ago
14

How to find the volume of a prism

Mathematics
2 answers:
Advocard [28]3 years ago
5 0

Answer:

the volume is V = 1120 inches squared

Step-by-step explanation:

GrogVix [38]3 years ago
5 0

Answer:

to find volume you just mutiply the length x width x height

Step-by-step explanation:

1120 = 20 x 7 x 8

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Suppose the radius of a circle is 8. What is its circumference? Enter either an exact answer or use pi or use 3.14 for pi and en
GrogVix [38]
8*3.14=  25.12 * 2= 
Circumference= 50.24
4 0
3 years ago
Read 2 more answers
If k(x) = 5x – 6, which expression is equivalent to (k + k)(4)?. . 5(4 + 4) – 6. 5(5(4) – 6) – 6. 54 – 6 + 54 – 6. 5(4) – 6 + 5(
Dmitrij [34]
K(x) = 5x-6
(k+k)(x) = k(x) + k(x) = 5x-6 +5x-6
just plug in x = 4,
(k+k)(4) = 5(4) -6 + 5(4) -6
thats your answer, the last option.
6 0
3 years ago
Read 2 more answers
Can someone help me with this please? Thanks! <br><br> Show your work :)
Alexandra [31]
I got 1/81. because u spun the spinner twice i multiplied the regions you are allowed to get by 2 which makes your denominator 18 then you can get the numbers 3 and 6 twice since the spinner is used twice then multiplied them together. so it would be 2/18 x 2/18 = 4/324 which equals 1/81
6 0
3 years ago
50 POINTS I WILL GIVE BRAINLIEST!!! HELPPPPPPPP....explain how to construct congruent line segments..
tigry1 [53]

Answer:

1. Construct a segment XY on the a sketch.

2. Elsewhere on the sketch, draw a line and create a point on the line . Label the point P.

3. Select point P and line segment XY.  Construct circle by center and radius.

4. Select the circle and line, then construct point of intersection. Label one of the points of intersection point Q.

5. Select point P and point Q.  Construct segment PQ.

6. Select the circle, line, and other point of intersection.  Hide objects.  Only segment XY and segment PQ should be visible on the sketch.

7. Try dragging point X.  Segment PQ should lengthen and shorten as segment XY does.

3 0
2 years ago
Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel
Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e (x_o, y_o, z_o) and are parallel to \dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

\begin  {vmatrix} x = 4+ t   \\ \\  \dfrac{x-4}{1} = t  \begin {vmatirx} \end {vmatrix}\begin {vmatrix} y = - 5+2t  \\ \\ \dfrac{y+5}{2}  =t      \end {vmatrix}\begin {vmatrix} z =2+t  \\ \\ \dfrac{z-2}{1}  =t      \end {vmatrix}

∴

\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}

8 0
3 years ago
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