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2 years ago

Please help i am confused

Mathematics

1 answer:

son4ous [18]2 years ago

7 0

Answer: (4)

Step-by-step explanation:

The parts listed in the congruence statements don't correspond, so they aren't necessarily congruent.

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( 3 1 ) 2 +3 2 =left parenthesis, start fraction, 1, divided by, 3, end fraction, right parenthesis, squared, plus, 3, squared

Fynjy0 [20]

Answer:

I guess that we have the equation:

(1/3)^2 + 3^2 =

And we want to solve it.

Here just remember that:

3^2 = 3*3 = 9

then:

(1/3)^2 = (1/3)*(1/3) = 1/9

replacing these in the expression, we get:

(1/3)^2 + 3^2 = 1/9 + 9

If we want to write this as a single fraction, we can rewrite:

1/9 + 9 = 1/9 + (9/9)*9

= 1/9 + 81/9 = (1 + 81)/9 = 82/9

4 0

3 years ago

HELP ASAP

almond37 [142]

Answer:

Step-by-step explanation:

let the number of dimes=x

and number of quarters=y

x=2y

0.10 x+0.25 y=2.25

or 10x+25y=225

divide by 5

2x+5y=45

2 ×2y+5y=45

4y+5y=45

9y=45

y=45/9=5

x=2y=2×5=10

Hence dimes=10

quarters=5

6 0

3 years ago

How could you solve for x in the problem 3x+7=-x-1

lukranit [14]

Hello there!

$3x+7=-x-1\\$

Explanation:

↓↓↓↓↓↓↓↓↓↓↓↓

First you had to subtract by 7 from both sides of the equation.

$3x+7-7=-x-1-7$

Simplify

$3x=-x-8$

Then you add by x from both sides of the equation.

$3x+x=-x-8+x$

Simplify

$4x=-8$

Divide by 4 from both sides of the equation.

$\frac{4x}{4}=\frac{-8}{4}$

Simplify it should be the correct answer.

$x=-2$

Answer⇒⇒⇒x=-2

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

7 0

3 years ago

Third-degree, with zeros of -4, -2, and 1, and a y-intercept of -11.

Cloud [144]

$\begin{cases} x = -4\implies &x+4=0\\ x=-2\implies &x+2=0\\ x=1\implies &x-1=0 \end{cases}\qquad \implies (x+4)(x+2)(x-1)=\stackrel{y}{0}$

now, that's the equation or polynomial in factored form, hmmm we also know that it has a y-intercept of -11, namely, when x = 0 y = -11, well let's plug in a factor to it, that will reflect those values, namely say hmmm factor "a", so

$(x+4)(x+2)(x-1)=y\qquad \stackrel{\textit{adding "a" factor for vertical shift}}{a(x+4)(x+2)(x-1)}=y \\\\\\ \stackrel{\textit{we know that when x = 0, y = -11}}{a(0+4)(0+2)(0-1)=-11}\implies -8a=-11\implies a=\cfrac{-11}{-8}\implies a = \cfrac{11}{8} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\mathbb{FOIL}}{\cfrac{11}{8}(x^2+6x+8)}(x-1)=y\implies \cfrac{11}{8}(x^3+6x^2+8x-x^2-6x-8)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{11}{8}(x^3+5x^2+2x-8)=y~\hfill$

4 0

2 years ago

What are matching partial products?

motikmotik

Just say partial products that relate to one another

5 0

3 years ago