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2 years ago

L –3 | A. 3 B. -3 C. 0.30 D. 30

Mathematics

1 answer:

Vesnalui [34]2 years ago

7 0

Answer:

| -3 | = 3

Step-by-step explanation:

minus is removed

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Daniel has $650 in his checking account. He would like to spend some of his money on a new Playstation 4 and a game. He wants to

d1i1m1o1n [39]

Answer:

The answer would be D t is less than or equal to 375

Step-by-step explanation:

Because he wants to save 275 and 275+375 is 650. so if he wants to save at least 275 he can only spend a max of 375.

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2 years ago

Pls solve this question

just olya [345]

Answer:

${( \sqrt{ {x}^{ - 3} }) }^{5} = ({( {x}^{ - 3}) }^{ \frac{1}{2} } ) ^{5} \\ \\ = {x}^{ - 3 \times \frac{1}{2} \times 5} = {x}^{ - \frac{15}{2} } = \frac{1}{ {x}^{ \frac{15}{2} } }$

<em><u>Or ;</u></em>

${( \sqrt{ {x}^{ - 3} } )}^{5} = {( \sqrt{ \frac{1}{ {x}^{3} }} })^{5} = ( { \frac{ \sqrt{1} }{ \sqrt{ {x}^{3} } } })^{5} \\ \\ = ( { \frac{1}{ \sqrt{x} \sqrt{ {x}^{2} } } })^{5} = ({ \frac{1}{ x\sqrt{x} }})^{5} \\ \\ = ( \frac{ {(1)}^{5} }{ {x}^{5} ( \sqrt{x} )^{5} } ) = \frac{1}{ {x}^{5} \times {x}^{ \frac{1}{2} \times 5 } } \\ \\ = \frac{1}{ {x}^{5} \times {x}^{ \frac{5}{2} } } = \frac{1}{ {x}^{5} \sqrt{ {x}^{5} } } \\ \\ = \frac{1}{ {x}^{5} \sqrt{ {x}^{4} {x}^{1} } } = \frac{1}{ {x}^{5} \sqrt{x} \sqrt{( { {x}^{2} })^{2} } } \\ \\ = \frac{1}{ {x}^{5} {x}^{2} \sqrt{x} } = \frac{1}{ {x}^{7} \sqrt{x} } = \frac{ \sqrt{x} }{ {x}^{8} }$

3 0

1 year ago

Consider rectangle WXYZ with diagonal XZ. Rectangle W X Y Z is cut diagonally into 2 equal triangles. Angle Y X Z is 26 degrees

shusha [124]

Answer:

26 i think

Step-by-step explanation:

next time try upload a picture so it for easier for us to answer

6 0

2 years ago

Read 2 more answers

PLZ I REALLY NEED HELP Find the values of x and y in parallelogram PQRS

brilliants [131]

Answer:

x = 5 and y = 7

Step-by-step explanation:

Given that,

PT = x+2, TR=y, QT=2x, TS=y+3

We need to find the value of x and y.

The diagonal of a parallelogram bisect each other. It means,

PT = TR and QT = TS

ATQ,

y = x+2 ...(1)

and

y+3 = 2x ....(2)

Subtracting equation (1) and (2).

y-(y+3) = x+2-2x

y-y-3 = 2-x

-3 = 2-x

x = 2+3

x = 5

Put the value of x in equation (1).

y = 5 + 2

y = 7

Hence, the values of x and y are x = 5 and y = 7.

4 0

2 years ago

Look at image please

sashaice [31]

Answer:

the answer is going to be letter b

8 0

1 year ago