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3 years ago

Does this have, one solution, no solutions, or infinite solutions?

Mathematics

2 answers:

MA_775_DIABLO [31]3 years ago

6 0

Answer:

Infinite

Step-by-step explanation:

Cuz i is a smart human

N76 [4]3 years ago

3 0

Answer:

The answer is <u>No Solution</u>

Step-by-step explanation:

You might be interested in

SHOW YOUR WORK! Marking the best response with brainliest:) thank u

sdas [7]

Step-by-step explanation:

Perimeter of a rectangle = 2l + 2w

where

l is the length

w is the width

From the question

The length is 3x + 7

The width is 2x + 2

The perimeter is

Simplify

6x + 7 + 4x + 4

Group like terms

6x + 4x + 7 + 4

We have the final answer as

Hope this helps you

6 0

3 years ago

What is a rational number between -1/4 and 2/5

kramer

1/5

Any number that can be written in fraction form is a rational number, so numbers like -1/8, 0, 1/4, etc. all work.

4 0

3 years ago

What interest will be charged on a $300 loan for 6 months if the rate is 18% simple interest?

Yakvenalex [24]

Answer: 540.00

Step-by-step explanation:

4 0

3 years ago

14/2 three equivalent ratios

Anna71 [15]

Answer:

Three equivalent ratios of $\frac{14}{2}$ are $\frac{7}{1}$ , $\frac{21}{3}$ and $\frac{28}{8}$ .

Step-by-step explanation:

We are given a fraction $\frac{14}{2}$ .

We need to find the three equivalent ratios/fractions of $\frac{14}{2}$ .

The common factor of 14 and 2 is 2.

So, dividing top and bottom by 2, we get

14÷2 = 7 and 2÷2 that is $\frac{7}{1}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

7 × 3 = 21

1 × 3 = 6

So, we get another equivalent fraction $\frac{21}{6}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

7 × 4 = 28

1 × 4 = 4

So, we get another equivalent fraction $\frac{28}{4}$ .

Therefore, three equivalent ratios of $\frac{14}{2}$ are $\frac{7}{1}$ , $\frac{21}{3}$ and $\frac{28}{4}$ .

6 0

3 years ago

Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.

Mkey [24]

Answer:

$1.~e^{-2} \\2.~e^{\frac{7}{2}}\\3.~e^8\\4.~e^{\frac{-11}{2}}$

Step-by-step explanation:

We have to simplify the given exponential exponents.

Exponential Properties:

$e^0 =1\\e^a.e^b = e^{a+b}\\\\\displaystyle\frac{e^a}{e^b} = e^{a-b}\\\\(e^a)^b = e^{ab}\\\\e^{-a} = \frac{1}{e^a}$

Simplification takes place in the following manner:

$(e^{-3})^\frac{2}{3}\\(e^a)^b = e^{ab}\\=e^{-3\times \frac{2}{3}}\\=e^{-2}$

$(e^4)(e^{\frac{-1}{2}})\\e^a.e^b = e^{a+b}\\=e^{(4+\frac{-1}{2})} \\= e^{\frac{7}{2}}$

$(e^{-2})^{-4}\\(e^a)^b = e^{ab}\\= e^{-2\times -4}\\=e^8$

$(e^{-4})(e^{\frac{-3}{2}})\\e^a.e^b = e^{a+b}\\=e^{(-4+\frac{-3}{2})} \\= e^{\frac{-11}{2}}$

4 0

3 years ago