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2 years ago

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Jerry drove 10 miles in half the time it took Marcy to drive 10 miles. If jerry average speed was r miles per hour what was Marc

y average speed in miles per hour

1 answer:

Diano4ka-milaya [45]2 years ago

8 0

Answer:

0.5r

Step-by-step explanation:

If Jerry took half the time Macy did then that means Macy took twice as long as Jerry. If Jerry’s speed was r then Macy’s speed is 0.5r or half the speed of Jerry. This will mean Macy goes half as slow and takes twice as long to get there.

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Interpret the meaning of the point (9, 378).

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Answer:

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and intersects at 378 on the y-axis

Step-by-step explanation:

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A customer at a store paid $64 for 3 large candles and 4 small candles. At the same store, a second customer paid $4 more than t

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system of equations can be used to find the price in dollars of each large candle, x, and each small candle, y is $x+8y=68$ and $3x+4y=64$ .

<u>Step-by-step explanation:</u>

Here we have , A customer at a store paid $64 for 3 large candles and 4 small candles. At the same store, a second customer paid $4 more than the first customer for 1 large candle and 8 small candles. The price of each large candle is the same, and the price of each small candle is the same. We need to find Which system of equations can be used to find the price in dollars of each large candle, x, and each small candle, y . Let's find out:

Let the price in dollars of each large candle, x, and each small candle, y .So

A customer at a store paid $64 for 3 large candles and 4 small candles

Equation is :

⇒ $3x+4y=64$ .....(1)

At the same store, a second customer paid $4 more than the first customer for 1 large candle and 8 small candles.

Equation is :

⇒ $x+8y=68$ .......(2)

3(2)-(1) i.e.

⇒ $3(x+8y)-(3x+4y)=68(3)-64$

⇒ $20y=140$

⇒ $y=7$

So , $x+8y=68$

⇒ $x+8(7)=68$

⇒ $x=12$

Therefore , system of equations can be used to find the price in dollars of each large candle, x, and each small candle, y is $x+8y=68$ and $3x+4y=64$ .

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What is the following product? RootIndex 3 StartRoot 16 x Superscript 7 Baseline EndRoot times RootIndex 3 StartRoot 12 x Supers

KengaRu [80]

Answer:

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }$

Step-by-step explanation:

Given

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9}$

Required

Find the products

From laws of indices;

$\sqrt[m]{a} * \sqrt[m]{b} = \sqrt[m]{a*b}$

So;

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16x^7 * 12x^9}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16* x^7 * 12 * x^9}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^7 * x^9}$

From laws of indices

$a^m * a^n = a^(m+n); So,$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{7+9}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{16}}$

Expand 16 * 12

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4*4*4*3* x^{16}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4^3 *3* x^{16}}$

From laws of imdices

$a^{\frac{1}{m}} = \sqrt[m]{a}$

So;

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^3 *3* x^{16}})^{\frac{1}{3}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^{3*{\frac{1}{3}}} *3^{{\frac{1}{3}}}* x^{16*{\frac{1}{3}}}})$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{\frac{16}{3}}}$

Divide 16 by 3 (Write as ,mixed number)

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5\frac{1}{3}}}$

Split mixed numbers

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5+\frac{1}{3}}}$

Apply law of indices

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5}*{x ^\frac{1}{3}}}$

Reorder

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 * x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}$

Apply law of indices

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3} *\sqrt[3]{x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3*x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }$

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2 years ago