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marishachu [46]
3 years ago
8

The sum of one and the product of four and a number x

Mathematics
2 answers:
frez [133]3 years ago
7 0
1+(4x) ??????????? I think so
patriot [66]3 years ago
4 0
Easy...  1+4x (x representing a number not times*)
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A 3 pound pork loin can be cut into 10 pork chops with equal weight. How many ounces is each pork chop?
Misha Larkins [42]

Answer

Find out the how many ounces is each pork chop .

To prove

Relationship between pound and ounce .

1 pound = 16 ounce

As given

A 3 pound pork loin can be cut into 10 pork chops with equal weight.

Now for 3 pound pork loin = 3 × 16

                                            = 48 ounces

now 48 ounces of  pork loin can be cut into 10 pork chops with equal weight.

Than

weight\ of\ each\ pork\ chops = \frac{48}{10}

                                                         = 4.8 ounces

Therefore the weight of each ounce be 4.8 ounces .





6 0
3 years ago
Find an equation of the circle having the given center and radius.<br> Center ( - 1, 1), radius 7
suter [353]

Answer:

(x+1)^2 + (y+1)^2 = 49

Step-by-step explanation:

We can write the equation of a circle as

(x-h)^2 + (y-k)^2 = r^2

where (h,k) is the center and r is the radius

(x- -1)^2 + (y- -1)^2 = 7^2

Simplifying

(x+1)^2 + (y+1)^2 = 49

7 0
3 years ago
Read 2 more answers
Add and Subtract Rational Expressions with a Common Denominator
Sladkaya [172]

Answer:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}

Operating on the numerators:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}

Removing parentheses:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}

Simplifying:

\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}

The expression cannot be further simplified.

7 0
3 years ago
Jennifer has a credit card with an APR of 10.22% and a billing cycle of 30 days. The following table shows Jennifer’s credit car
Nimfa-mama [501]

Answer:

The answer is c.

Step-by-step explanation:

5 0
3 years ago
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If 2 is added to both the numerator and denominator of a fraction its value becomes half. If two is subtracted from both the num
Hitman42 [59]

Answer:

The original fraction is \frac{6}{14}.

Step-by-step explanation:

Solution,

Let the fraction be'\frac{x}{y}'.

Where 'x' is the numerator and 'y' is the denominator.

Now according to question, if 2 is added to both the numerator and denominator the value of the fraction becomes 1/2.

So, framing the above sentence in equation form, we get;

\frac{x+2}{y+2}=\frac{1}{2}

On using cross multiplication method, we get;

2(x+2)=y+2\\\\2x+4=y+2\\\\2x+4-2=y\\\\\therefore\ y=2x+2\ \ \ \ equation\ 1

Now according to question, if 2 is Subtracted to both the numerator and denominator the value of the fraction becomes 1/3.

So, framing the above sentence in equation form, we get;

\frac{x-2}{y-2}=\frac{1}{3}

On using cross multiplication method, we get;

3(x-2)=y-2\\\\3x-6=y-2\\\\3x-y=-2+6\\\\3x-y=4 \ \ \ \ equation\ 2

Now Substituting the equation 1 in equation 1 we get;

3x-y=4\\\\3x-(2x+2)=4\\\\3x-2x-2=4\\\\3x-2x=4+2\\\\x=6

Now Substituting the value of x in equation 1 we get;

y=2x+2\\\\y=2\times6+2 = 14

Hence The original fraction is \frac{6}{14}.

4 0
3 years ago
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