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3 years ago

Jared has four pieces of rope that are each 12.75 feet long. What is the total length of rope he has

Mathematics

2 answers:

Nostrana [21]3 years ago

4 0

Jared's rope is 51 feet long

TiliK225 [7]3 years ago

3 0

He has 51 feet of rope

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Which statement is true

dybincka [34]

Jill will make less money if they work for 4 hours

3 0

3 years ago

Please help me only say something if you really know the answer

Diano4ka-milaya [45]

Answer:

Step-by-step explanation:

Starting with 7,000, after 0 years there will be no increase so you still have 7,000.

The fist year you increase by 5% of 7,000.

.05x7000=350

You have a 350 increase, add that to the original 7000 to find the actual population after 1 year (domain value 1).

After 1 year: 7350

For year 2 there is an increase of 5% again, only this time we find 5% of 7350 since that was the previous years population.

.05x7350=368

Add that to previous population.

368+7350=7718

At this point so far the yearly populations have been (7000, 7350, 7718)

Answer choice B is the only one to have this progression.

3 0

3 years ago

4. The base of a triangle is 4 3/4ft. The height of the

astra-53 [7]

Answer:

39 $\frac{9}{10\\}$

Step-by-step explanation:

4 $\frac{3}{4}$ × 8 $\frac{2}{5}$ = 39 $\frac{9}{10\\}$

6 0

3 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

Please help!!

postnew [5]

$2 \frac{5}{8} \div 2 \frac{1}{2}$
Change the mixed fractions to improper fractions.
To do this, multiply the whole number part by the denominator. Add that to the numerator. Then write the result on top of the denominator.

You would now have:
$\frac{21}{8} \div \frac{5}{2}$
Change the division sign to a multiplication sign by turning the second fraction upside down.
That is,
$\frac{21}{8} \times \frac{2}{5}$

The answer is:
$\frac{42}{40}$

This can be simplified to:
$\frac{21}{20}$

8 0

3 years ago