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3 years ago

If it takes 15 gallons of gas to drive 330 miles how many miles can be driven using 20 gallon of gas

1 answer:

3 0

15 / 330 = 20 / x...15 gal. to 330 miles = 20 gal to x miles
cross multiply because this is a proportion
(15)(x) = (20)(330)
15x = 6600
x = 6600/15
x = 440 <=== 20 gallons will take u 440 miles

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Zeke read z books over the summer. Michelle read twice as many books over the summer as Zeke. Tay read 3 more books over the sum

Vikki [24]

Answer: 4z + 3

Step-by-step explanation:

Number of books read by Zeke = z

Michelle read twice as many book as Zeke. This will be: = 2 × z = 2z

Tay read 3 more books than Zeke. This will be: = z + 3

The expression that represents the total number of books read over the summer by Zeke, Michelle and Tay would be the addition of the books read by each person. This will be:

= z + 2z + z + 3

= 4z + 3

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3 years ago

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musickatia [10]

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Step-by-step explanation:

Just solving the two equations

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y = 4/3 x +3

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by substracting 2 from 1

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3 years ago

Find the product of (3x + 7y)(3x - 7y). (2 points) PLEASE I NEED THIS NOW

Murljashka [212]

9x^2 - 49y^2

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3 years ago

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If alpha and beta are the angles in the first quadrant tan alpha = 1/7 and sin beta =1/ root 10 then usind the formula sin (A +B

zysi [14]

Answer:

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

Step-by-step explanation:

$\alpha \text{ and } \beta \text{ in Quadrant I}$

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<u>Using Pythagorean Identities</u>:

$\boxed{\sin^2(\theta)+\cos^2(\theta)=1} \text{ and } \boxed{1+\tan^2(\theta)=\sec^2(\theta)}$

$\left(\frac{\sqrt{10} }{10} \right)^2+\cos^2(\beta)=1 \Longrightarrow \cos(\beta)=\sqrt{1-\frac{10}{100}} =\sqrt{\frac{90}{100}}=\frac{3\sqrt{10}}{10}$

$\text{Note: } \cos(\beta) \text{ is positive because the angle is in the first qudrant}$

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$\Longrightarrow \frac{49}{50}=\cos^2(\alpha) \Longrightarrow \cos(\alpha)=\sqrt{\frac{49}{50} } =\frac{7\sqrt{2}}{10}$

$\text{Now let's find }\sin(\alpha)$

$\sin^2(\alpha)+\left(\frac{7\sqrt{2} }{10}\right)^2=1 \Longrightarrow \sin^2(\alpha) +\frac{49}{50}=1 \Longrightarrow \sin(\alpha)=\sqrt{1-\frac{49}{50}} = \frac{\sqrt{2}}{10}$