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Arte-miy333 [17]

2 years ago

12

Pears cost $0.92 per pound and apples cost $1.10 per pound. Mr.Bonilla bought 3.75 pounds of pears and 2.1 pounds of apples. How

much did he pay for the pears and the apples?

1 answer:

hodyreva [135]2 years ago

7 0

Mr. Bonilla payed $3.45 for the pears and $2.31 for the apples (.92 x 3.75 and 2.1 x 1.10)

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Evaluate the expression if a=2,b=-3,C=-1, and D=4<br><br><br> -2(b^2-5c)

Alenkasestr [34]

$\qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star$

Equivalent value = -28

$\textsf{ \underline{\underline{Steps to solve the problem} }:}$

$\qquad❖ \: \sf \: - 2( {b}^{2} - 5c)$

( put the values )

$\qquad❖ \: \sf \: - 2 \{( - 3) {}^{2} - 5( - 1) \}$

$\qquad❖ \: \sf \: - 2(9 - (- 5))$

$\qquad❖ \: \sf \: - 2(9 + 5)$

$\qquad❖ \: \sf \: - 2 \times 14$

$\qquad❖ \: \sf \: - 28$

$\qquad \large \sf {Conclusion} :$

-2(b² - 5c) = -28

7 0

1 year ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

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8th grade math help please zoom in on the question for a better look (will give brainliest and thanks if it's right)

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