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gregori [183]
3 years ago
13

Chloe has 26 1/4 yards of ribbon to wrap presents. She needs 3 3/4 yards for each present. How many presents can Chloe wrap?

Mathematics
1 answer:
Mashcka [7]3 years ago
3 0

Chloe could wrap 7 presents

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3 years ago
If B = 0 in a linear equation of the form Ax + By = C, then which statement is also true? A. The graph of this equation is a hor
Vika [28.1K]

Answer:

<h3><u>A) Its horizontal line </u></h3>

Step-by-step explanation:

Ax + By = C

so Ax + (0)y = C

a= c /x

its horizontal

5 0
3 years ago
What is an expression for vector x involving vector r, vector s, and vector t?
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X = -3r + s - t ....
3 0
4 years ago
Please help with question 20! Thanks&lt;3
kkurt [141]

Answer:

Population of Town A = 19,575

Population of Town B = 52,200

Population of Town C = 20,880

Step-by-step explanation:

The ratio of..

A : B = 3 : 8

B : C = 5 : 2

Both A : B and B : C have B in their ratios. What we can do is make B the same in both ratios and then create a ratio between all three towns.

A : B = (3 : 8) x 5 = 15 : 40

B : C = (5 : 2) x 8 = 40 : 16

A : B : C

15 : 40 : 16

Now, to find the population of each town we can write the following equation using the numbers from our ratio between all three towns and set it to the given population.

15x + 40x + 16x = 92,655

Solve for x.

71x = 92,655

x = 1,305

Finally, we can use the value of x to find the population of each town.

Population of Town A = 15x

Population of Town A = 15(1,305)

Population of Town A = 19,575

Population of Town B = 40x

Population of Town B = 40(1,305)

Population of Town B = 52,200

Population of Town C = 16x

Population of Town C = 16(1,305)

Population of Town C = 20,880

~Hope this helps!~

3 0
4 years ago
Solving Rational equations. LCD method. Show work. Image attached.
posledela

(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12

So in order to get all the fractions to have a common denominator, we need to multiply \dfrac k{k-2} by \dfrac{k-6}{k-6}, and \dfrac1{k-6} by \dfrac{k-2}{k-2}:

\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}

\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}

Now,

\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}

As long as k\neq2 and k\neq6 (which we can't have because otherwise k^2-8k+12=0), we can cancel k^2-8k+12 in the denominators on both sides:

4=(k^2-6k)+(k-2)

4=k^2-5k-2

0=k^2-5k-6

We can factorize the right side:

0=(k-6)(k+1)

which tells us that k=6 and k=-1 are solutions.

4 0
4 years ago
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