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xz_007 [3.2K]
4 years ago
12

Need help ASAP!! thank you sorry if u can’t see it good :(

Mathematics
1 answer:
Wittaler [7]4 years ago
5 0

Answer/Step-by-step explanation:

==>Given:

=>Rectangular Pyramid:

L = 5mm

W = 3mm

H = 4mm

=>Rectangular prism:

L = 5mm

W = 3mm

H = 4mm

==>Required:

a. Volume of pyramid:

Formula for calculating volume of a rectangular pyramid us given as L*W*H

V = 5*3*4

V = 60 mm³

b. Volume of prism = ⅓*L*W*H

thus,

Volume of rectangular prism given = ⅓*5*3*4

= ⅓*60

= 20mm³

c. Volume of the prism = ⅓ x volume of the pyramid

Thus, 20 = ⅓ × 60

As we can observe from our calculation of the solid shapes given, the equation written above is true for all rectangular prism and rectangular pyramid of the same length, width and height.

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puteri [66]

I believe it is $0.50.

$2 divided by 4 equals $0.50.

8 0
3 years ago
The perimeter of an equilateral triangle is at most 45centimeters. Find the possible length of each side.
MAXImum [283]
Imdomt really understand but i done 180 divide 3 = 60
4 0
3 years ago
In the figure below, lines L and M are parallel and line AB = line BC = line AC. What is the measure of the exterior angle
Kryger [21]

Answer:

120º

Step-by-step explanation:

angles of an equilateral triangle are equal so..

180 ÷ 3 = 60º

to get the exterior m⦣C

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6 0
3 years ago
Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first
BaLLatris [955]

Answer:

\cos(x+y) goes with -\frac{\sqrt{6}+\sqrt{2}}{4}

\sin(x+y) goes with \frac{\sqrt{6}-\sqrt{2}}{4}

\tan(x+y) goes with \sqrt{3}-2

Step-by-step explanation:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

We are given:

\sin(x)=\frac{\sqrt{2}}{2} which if we look at the unit circle we should see

\cos(x)=\frac{\sqrt{2}}{2}.

We are also given:

\cos(y)=\frac{-1}{2} which if we look the unit circle we should see

\sin(y)=\frac{\sqrt{3}}{2}.

Apply both of these given to:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}

\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}

\frac{-\sqrt{2}-\sqrt{6}}{4}

-\frac{\sqrt{6}+\sqrt{2}}{4}

Apply both of the givens to:

\sin(x+y)

\sin(x)\cos(y)+\sin(y)\cos(x) by addition identity for sine.

\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}

\frac{-\sqrt{2}+\sqrt{6}}{4}

\frac{\sqrt{6}-\sqrt{2}}{4}

Now I'm going to apply what 2 things we got previously to:

\tan(x+y)

\frac{\sin(x+y)}{\cos(x+y)} by quotient identity for tangent

\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}

-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}

-\frac{8-2\sqrt{12}}{4}

There is a perfect square in 12, 4.

-\frac{8-2\sqrt{4}\sqrt{3}}{4}

-\frac{8-2(2)\sqrt{3}}{4}

-\frac{8-4\sqrt{3}}{4}

Divide top and bottom by 4 to reduce fraction:

-\frac{2-\sqrt{3}}{1}

-(2-\sqrt{3})

Distribute:

\sqrt{3}-2

6 0
3 years ago
Add.<br><br> Express your answer in simplest form. <br><br><br> 5/10+3/10
AlexFokin [52]
Rules in adding fractions:

1) Make sure that they have common denominator.
5/10 + 3/10

2) Since both fraction have a denominator of 10, add the numerator and place the sum above the denominator

5/10 + 3/10 = (5 + 3)/10 = 8/10

3) Simplify the fraction
8/10 can still be simplified into 4/5

8 ÷ 2 = 4
10 ÷ 2 = 5

5/10 + 3/10 = 4/5
8 0
3 years ago
Read 2 more answers
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