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WINSTONCH [101]
3 years ago
8

Richard has three times as many crayons as Jordan. Together they have 60 crayons. How many crayons does

Mathematics
1 answer:
vesna_86 [32]3 years ago
5 0

Answer: 44 Crayons

Step-by-step explanation:

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Plsssss help due today !!! ill give brainliest
Grace [21]

Answer:

1.5

Step-by-step explanation:

We can set up the equation 0.50X = 0.750

Divide by 0.5 and we get .75/.5=1.5

4 0
3 years ago
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Please help if your awake lol ugh it’s a question check the picture :)
ladessa [460]
I really hope that helps cause I went all out and best of wishes!!

8 0
2 years ago
Andrew bashir and candy are trying to save money for a birthday party.If andrew saves 1/4 of the total needed bashir saves 2/5 a
DerKrebs [107]

Answer:

1/4.

Step-by-step explanation:

That is 1 - (1/4 + 2/5 + 1/10).

The Lowest common multiple  4 5 and 10 is 20 so we have

20/20 - (5/20 + 8/20 + 2/20)

=  20/20 - 15/20

= 5/20

= 1/4 answer.

5 0
3 years ago
Which of the following is a linear function?
Rama09 [41]
\boxed{ \textbf{ \huge{Answer: A}}}

A linear function has 1 as the highest power of the variable.

A. f(x) = 2 - 7x

here, the highest power of the variable x is 1.
Hence it is a LINEAR function.
___________

B. f(x) = 2 + x + x^2

here, the highest power of the variable x is 2.
Hence it is NOT a linear function.
_______________

c.f(x) = \sqrt{2 + 7x}
here, the highest power of the variable x is 1/2.
Hence it is NOT a linear function.
________________
6 0
2 years ago
Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al
MissTica

\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5

\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}

4 0
2 years ago
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