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3 years ago

Tell whether each equation has one ,zero,or infinitely many solutions

Mathematics

1 answer:

zubka84 [21]3 years ago

6 0

The answer to the question

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Which models show 60%? Check all that apply.

goldfiish [28.3K]

Answer:

OPTION A AND B.

AS OPTION A IS 3/5=6/10=60%

Option C=60/100=60%

6 0

3 years ago

Read 3 more answers

Find all values of k for which the equation 3x^2−4x+k=0 has no solutions.

dusya [7]

The given equation has no solution when K is any real number and k>12

We have given that

3x^2−4x+k=0

△=b^2−4ac=k^2−4(3)(12)=k^2−144.

<h3>What is the condition for a solution?</h3>

If Δ=0, it has 1 real solution,

Δ<0 it has no real solution,

Δ>0 it has 2 real solutions.

We get,

Δ=k^2−144 here Δ is not zero.

It is either >0 or <0

Δ<0 it has no real solution,

Therefore the given equation has no solution when K is any real number.

To learn more about the solution visit:

brainly.com/question/1397278

5 0

3 years ago

Read 2 more answers

Please help me with these and please have explanations

NeTakaya

Answer:

1-B

2-C

3-D

Step-by-step explanation:

In the attached file

6 0

3 years ago

Which fraction has the value that's equal to 3/4

mamaluj [8]

One of the fractions that’s equal to $\frac{3}{4}$ is $\frac{12}{16}$

<u>Solution:</u>

Given that , we have to find fractions which has the same value as that of the fraction $\frac{3}{4}$

Now, we know that, there are several fractions with values equal to $\frac{3}{4}$

To find them, just multiply the numerator and denominator by the same number.

$\begin{array}{l}{3 \times 2=6} \\\\ {4 \times 2=8}\end{array}$

Therefore, $\frac{6}{8}$ is equal to $\frac{3}{4}$

We can do the same with 4, to get $\frac{12}{16}$ , or any other number beyond that.

Hence, one of the fractions that’s equal to $\frac{3}{4}$ is $\frac{12}{16}$

4 0

3 years ago

Please help me find the area of the triangle in terms of x!

Vitek1552 [10]

Answer:

$Solution,\\Area~of~triangle=\frac{1}{2} (base)(height) = \frac{1}{2}(2x+\frac{1}{4}) (x-8)\\~~~~~~~~~~~~~~~~~~~~~~~~ =\frac{1}{2}[2x^2-16x+\frac{1}{4}x-\frac{1}{4}(8)]\\ ~~~~~~~~~~~~~~~~~~~~~~~~ =\frac{1}{2}[2x^2-\frac{63}{4}x - 2]\\~~~~~~~~~~~~~~~~~~~~~~~~=x^2-\frac{63}{8}x-1$

7 0

3 years ago