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QveST [7]
3 years ago
9

PLEASE CAN SOMEONE EXPLAIN THIS TO ME . I DONT KNOW HOW TO DO IT

Mathematics
1 answer:
Gekata [30.6K]3 years ago
3 0

Answer:

35

Step-by-step explanation:

-5{3-2[1-4(3-2^2)]}

we work inside out

using PEMDAS

parenthesis, exponents, multiply and divide, add and subtract

the very most inside

(3-2^2)

do the exponents

3 - 4 = -1

replace where it was in the original equation

-5{3-2[1-4(-1)]}

take the next set of parenthesis (bracket are the same as parenthesis)

[1-4(-1)]

multiply

4*-1 = -4

1--4 = 1+4 = 5

replace  where it was in the original equation

-5{3-2[5]}

next set of parenthesis

{3-2[5]}

3 - 10 = -7

replace  where it was in the original equation

-5 {-7}

35

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ivolga24 [154]

Answer:

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Step-by-step explanation:

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8 0
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What are 2 binomials that are factors of this trinomial? x^2-x-20
liberstina [14]

Answer:

(x-5) (x+4)

Step-by-step explanation:

x^2 -x -20

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8 0
3 years ago
Read 2 more answers
Mrs. Valdez wanted to determine the number of people, p, she could safely take in her car to the fun run. She determined that p
dolphi86 [110]

Answer:

for the first question b,


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for the second question I need to see the whole inequality, generally there is a variable, and all i see is <-1, which is invalid.


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Step-by-step explanation:


5 0
3 years ago
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

3 0
4 years ago
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