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3 years ago

PLEASE PLEASEE HELP ITS PAST DUE AND ITS A TEST ILL GIVE BRAINLIEST

Mathematics

2 answers:

agasfer [191]3 years ago

5 0

Answer:

The length will be letter C, 50 m

Akimi4 [234]3 years ago

4 0

The length is c, 50 m

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prisoha [69]

Answer:

x>-3 is the answer hope it helps

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4 years ago

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Devon had 3/5 of a candy bar. Gina ate ½ of it. How much of it remains?

xz_007 [3.2K]

Answer:

3/10

Step-by-step explanation:

To find how much remains, you need to find 1/2 of 3/5 which is 1/2 * 3/5. To multiply the two fractions, multiply the two numerators and multiply the two denominators to get 3/10. Thus, 3/10 of the candy bar remains.

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3 years ago

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A car salesman had $65,100 in sales. He earned $1,953 in commission. What percent commission did he earn. Explain your answer.

ahrayia [7]

Answer:

1. The percent commission earned is 3%.

2. The loan period is 3.29 years.

Step-by-step explanation:

1. Salesman has $65,100 in sales. He earned $1,953 in commission.

Let the percent commission earn be x%

Therefore, x% of the sales equals $1,953

$\frac{x}{100} of $65,100 = $1,953$

$\frac{x}{100} * 65100 = 1953\\\frac{65100x}{100} = 1953$

We cross multiply

$\frac{65100x}{100} = 1953\\65100x = 1953 * 100\\65100x = 195300$

Divide both side by the coefficient of 'x' (65100)

$\frac{65100x}{65100} = \frac{195300}{65100} \\x = 3$

Therefore, the percent commission earned is 3%

2. Interest (I) = $299 Principal (P) = $1300 Rate (R) = 7%

The formula for finding interest is given as: $I = \frac{PRT}{100}$

Therefore, substituting into the formula, we have:

$299 = \frac{1300 * 7 * T}{100}$

We are finding the time it takes the loan to earn an interest of $299

$299 = \frac{1300 * 7 * T}{100} \\299 = \frac{9100T}{100}$

We cross-multiply:

$299 * 100 = 9100T\\29900 = 9100T$

Divide both side by the coefficient of T (9100)

$\frac{29900}{9100} = \frac{9100T}{9100}\\T = 3.29$

Therefore, the time taken for the loan to earn such interest is approximately 3.29 years

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4 years ago

Find the derivative of f(x) = 6x + 2 at x = 1.

earnstyle [38]

$f'(x_0)=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=6x+2;\ x_0=1\\\\f'(1)=\lim\limits_{x\to x_0}\dfrac{(6x+2)-(6\cdot1+2)}{x-1}=\lim\limits_{x\to x_0}\dfrac{6x+2-8}{x-1}\\\\=\lim\limits_{x\to x_0}\dfrac{6x-6}{x-1}=\lim\limits_{x\to x_0}\dfrac{6(x-1)}{x-1}=\lim\limits_{x\to x_0}6=6\\\\\\Answer:\boxed{f'(1)=6}$

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3 years ago

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A nationwide survey claimed that at least 65% of parents with young children condone spanking their child as a regular form of p

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Answer:

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