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3 years ago

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:

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}

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ivolga24 [154]

Answer:

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Step-by-step explanation:

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3 years ago

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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-5*4 = -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,

it doesn't make sense for choice a to have -2 people in the car. You can't have a negative amount of people. For choice choice c, you can't have half a person, and for choice d, Ms. Valdez can only have under 5 people, not five people exactly.

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.

hope this helps, if it does like it pls!!

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

Step-by-step explanation:

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

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

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

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

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

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

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

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

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4 years ago