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ruslelena [56]
3 years ago
10

Reiki receives a package in the shape of a cube. Each side of the package is 4 feet long.

Mathematics
2 answers:
Troyanec [42]3 years ago
8 0

Answer:

96 SQUARE FEET

Step-by-step explanation:I

LIEIK DUH, I took da test,coronavirus iz making me dumb, ngl, I'm losing braincells.

Volgvan3 years ago
3 0
6• (4)^2= 96
The answer is 96 square feet
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What is the number of ways to arrange 6 objects from a set of 10 different objects
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<span>10!/(10-6)! =
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Point E is the midpoint of AB and point F is the midpoint
GREYUIT [131]

The options are;

1) AB is bisected by CD

2) CD is bisected by AB

3) AE = 1/2 AB

4) EF = 1/2 ED

5) FD= EB

6) CE + EF = FD

Answer:

Options 1, 3 & 6 are correct

Step-by-step explanation:

We are told that Point E is the midpoint of AB. Thus, any line that passes through point E will bisect AB into two equal parts.

The only line passing through point E is line CD.

Thus, we can say that line AB is bisected by pine CD. - - - (1)

Also, since E is midpoint of Line AB, it means that;

AE = EB

Thus, AE = EB = ½AB - - - (2)

Also, we are told that F is the mid-point of CD.

Thus;

CF = FD

Point E lies between C and F.

Thus;

CE + EF = CF

Since CF =FD

Thus;

CE + EF = FD - - - (3)

5 0
3 years ago
I need help with finding interior/exterior angles. solve for “?”
Ivenika [448]

Answer:

3. 160

4. 105

5. 95

6. 30

7. 86

8. 130

All triangles equal 180 so when you add the other 2 degrees in the triangle then subtract that by 180 then you get the third angle. The a straight line also equals 180 degrees so when you get the third angle subtract it by 180 then you get the answer to the "?"

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1+secA/sec A = sin^2 A / 1-cos A​
Fofino [41]

Answer:  see proof below

<u>Step-by-step explanation:</u>

\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}

\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}

\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}

\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}

\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}

\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark

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