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3 years ago

9

At a restaurant, brunch cost $18.50. As the cost of ingredients rises, the cost of brunch also rises by 40%. What is the new pri

ce of brunch?

1 answer:

statuscvo [17]3 years ago

6 0

$25.90 because if you move the decimal over one place in 18.50 you get 1.85 that is 10% so you take that times four and get $7.50 add that to 18.50 and you get the answer

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You deposit $300 in a savings account. The account earns 2% simple interest per year.

Sholpan [36]

Answer:

Answer:

a. ----> $10

b. ----> $110

Step-by-step explanation:

Step-by-step explanation:

principle ( p ) = $100

time ( t ) = 10 years

rate ( r ) = 11%

simple interest = (p × r × t)÷ 100

= ( $100 × 11 × 10 )÷ 100

= 11000 ÷ 100

= $110

interest = simple interest - principle

interest = $110 - $100

= $10

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2 years ago

Which of the following is the equation of a line perpendicular to the line

Taya2010 [7]

Answer:

A. - 2x + 3y = 21

Step-by-step explanation:

The perpendicular lines have opposite reciprocal slopes.

<u>The given line has a slope of:</u>

m = -3/2

<u>The perpendicular slope is:</u>

m = 2/3

<u>And its y-intercept is:</u>

9 = 2/3*3 + b
b = 7

<u>The line is:</u>

y = 2/3x + 7

<u>Convert this into standard form and compare with answer choices:</u>

3y = 2x + 21
- 2x + 3y = 21

Correct choice is A

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2 years ago

Select ALL the expressions that are equivalent to -4(8x-3)

Vinvika [58]

Answer:

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2 years ago

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3 years ago

Read 2 more answers

Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

3 years ago