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3 years ago

Please help me!! I’m really bad at math

Mathematics

1 answer:

viktelen [127]3 years ago

6 0

10 squared plus 18 squared equals C squared is the set up
After solving you should get 20.59

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How would I graph this?

elena55 [62]

$y\leq 2x-5$

↝ x-intercept

$2x-5=0\\2x=5\\x=\frac{5}{2}$

↝ y-intercept

y = -5

↝ $y\leq 2x-5$ means that the shade/area is below the graph as shown in the first picture.

$y\leq -\frac{1}{2}x-3$

↝ x-intercept

$-\frac{1}{2}x-3=0\\-x-6=0\\-x=6\\x=-6$

↝ y-intercept

y = -3

↝ Since it's "≤" therefore. The shade/area is below the graph. For second picture.

First picture is first inequality

Second picture is second inequality.

3 0

3 years ago

Find the value of s if 57+s=68

olga2289 [7]

Answer

s = 11

Step-by-step explanation:

Find the value of S:

57 + s = 68

Substract 57 from both sides of the equation

s = 68 - 57

s = 11

6 0

3 years ago

Read 2 more answers

What is the name for a percent that is given or paid in addition to a service?

Vilka [71]

Answer:

its B

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the amount erica owes at the end of 6 years if $60000 is loaned to her at a rate of 4% compounded monthly

RSB [31]

Answer: $76,244.51

Step-by-step explanation:

You need to use the compound interest formula here.

First of all however, you need to convert the terms to monthly figures because the interest is compounded monthly.

4% in months = 4 / 12 = 4/12%

6 years = 6 * 12 = 72 months

Now use the compound interest formula:

= Amount * (1 + rate) ^ number of years

= 60,000 * ( 1 + 4/12%) ⁷²

= $76,244.51

4 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago