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irakobra [83]
3 years ago
12

Functions f(x) and g(x) are shown below.

Mathematics
1 answer:
Natali5045456 [20]3 years ago
7 0
To get the 8x part, shift left by 8. To get the 16 part, shift up by 16.
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Joanna brought a new purse for $65 shoes for $80 and a dress for $125. She has paid for mother back 2/3 of the cost of the items
Vikki [24]

Answer:

$90

Step-by-step explanation:

The total of all the items is $270. You are lucky that this is a multiple. Solve it like you were dividing 27 by 9, just add a 0. Good luck! - Justin <3

4 0
3 years ago
Josiaanswer this is it josia the question
makkiz [27]

Answer:

6? i am soooo confused

Step-by-step explanation:

3 0
2 years ago
ASAP
Gala2k [10]
35 minutes yes tag it’s let me know if I’m wrong
8 0
3 years ago
Five cupcakes and two cookies cost $19.75. Two cupcakes and four cookies cost $ 17.50.
alina1380 [7]

Answer:

Step-by-step explanation:

Let x and y represent the cost of a cupcake and cookie respectively.

Given that;

Five cupcakes and two cookies cost  $19.75.

5x+2y=19.75 ---------- 1

Two cupcakes  and four cookies cost $17.50.

2x+4y=17.50 --------------2

Let's solve the simultaneous equation by elimination;

Multiply equation 1 by 2;

10x+4y=39.50-------3\\

Subtract equation 2 from equation 3;

10x-2x+4y-4y=39.50-8x=22

8x=22

divide both sides by 8

\frac{8x}{8} =\frac{22}{8} \\x=2.75

Since we have the value of x, let substitute into equation 1 to get y;

5x+2y=19.75\\5(2.75)+2y=19.75\\13.75+2y=19.75\\2y=19.75-13.75\\2y=6\\y=\frac{6}{2} \\y=3.00

therefore , the cost of cupcakes and cookies are;

cupcakes=  2.75\\cookies= 3.00

PLEASE MARK ME AS BRAINLIEST

6 0
3 years ago
How to find upper and lower bound of a definite integral given an unknown equation?
worty [1.4K]
Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of  definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this 

\int\limits^6_1 t^{2} - 6t + 11dt 

So, i integrate this, 
( \frac{t^{3} }{3} - 3t^{2} + 11t) \int\limits^6_1

I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
4 0
3 years ago
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