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3 years ago

Can someone help me please

Mathematics

2 answers:

ad-work [718]3 years ago

5 0

Answer:

29 3/4

Step-by-step explanation:

60 1/4 + x=90

Lorico [155]3 years ago

4 0

29 3/4 i am pretty sure !

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Is -10/3 greater or less that the square root of -9

svp [43]

Step-by-step explanation:

The square root of -9 is greater than that

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3 years ago

Hello!! Can someone help me solve the equation 5.15x + 6 = 72.94 (I’m solving for x)

n200080 [17]

Answer:

A x = 13.00

Step-by-step explanation:

5.15x + 6 = 72.94

5.15x = 66.94

x = 12.998

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3 years ago

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Elan Coil [88]

No 3 should be inch 6 should be feet

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3 years ago

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8. Marvin's son is 3 ft tall. He casts a shadow of 1 foot. Marvin cast a

taurus [48]

Answer:

C. 6 ft

Step-by-step explanation:

If Marvin's son is 3 feet tall with a 1 foot shadow, Marvin would be 6 feet tall with a 2 foot shadow.

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3 years ago

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Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val

zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

$\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill$

$\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}$

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}$

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3 years ago