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4 years ago

10

a restaurant holds an anniversary celebration of 90 diners. Only 27 are offered free dessert, what is the percent of dinners not

offered free dessert?

1 answer:

alex41 [277]4 years ago

6 0

Answer:

70%

Step-by-step explanation:

(90-27)/90

=63/90

=0.7

Changing it to percentage:

=70%

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Solve the following inequality algebraically. <img src="https://tex.z-dn.net/?f=%20%7Cx%20-%201%7C%20%20%5Cleqslant%2013" id="Te

olya-2409 [2.1K]

lx-1 l ≤ 13

There are 2 solutions:

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and

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Solve each one

x-1 ≤ 13

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2 years ago

BRAINLYEST UPON RIGHT ANSWER AND 100 POINTS. This will test your knowledge.

sertanlavr [38]

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3 years ago

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2 years ago

Which of these is equivalent to p=2l+2w

vlabodo [156]

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3 0

3 years ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution<br> Need ASAP

aniked [119]

Answer:

$x_1=\frac{1+\sqrt{33} }{2},y_1=\frac{5+\sqrt{33} }{2}\\ \\ x_2=\frac{1-\sqrt{33} }{2},y_2=\frac{5-\sqrt{33} }{2}$

Explanation:

You must find the solution of the system using the substitution method.

System:

$y=-x^2+2x+10\\\\ y=x+2$

Substitute x+2 for y:

$x+2=y=-x^2+2x+10$

Pass all the terms to one side:

$x^2-2x+x+2-10\\ \\ x^2-x-8=0$

Use the quadratic formula to solve:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\ \\ x=\frac{-(-1)+/-\sqrt{(-1)^2-4(1)(-8)} }{2(1)}\\ \\ x=\frac{1+/-\sqrt{1+32} }{2}$

$x_1=\frac{1+\sqrt{33} }{2}\\ \\x_2=\frac{1-\sqrt{33} }{2}$

Use y = x + 2 to find the value of y:

$y_1=2+x_1=2+\frac{1+\sqrt{33} }{2}\\ \\y_1=\frac{5+\sqrt{33} }{2} \\\\y_2=2+x_2=2+\frac{1-\sqrt{33} }{2}\\ \\ y_2=\frac{5-\sqrt{33} }{2}$

6 0

3 years ago