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Pavel [41]
3 years ago
13

−z/4=−3/4 Helppppppp

Mathematics
1 answer:
stealth61 [152]3 years ago
6 0

Answer:

z=3

Step-by-step explanation:

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Answer this question please
ioda
The 2nd one. The data matches that on the Venn Diagram.

6 0
3 years ago
Which graph represents the function f(x)= 0.2^x + 3
Dennis_Churaev [7]

The third graph or bottom left graph represents f(x) = 0.2^{x} + 3.

Step-by-step explanation:

Step 1:

To determine which of the given graphs represents the equation f(x) = 0.2^{x} + 3, we substitute some values in the place of x.

When x=0, f(x) = 0.2^{x} + 3, f(0) = 0.2^{0} + 3= 4.

Anything with an exponent of 0 will equal 1.

So the graphs on the right side cannot be the answers.

Step 2:

Now we substitute another value to determine which graph represents f(x) = 0.2^{x} + 3.

When x=1, f(x) = 0.2^{x} + 3, f(1) = 0.2^{1} + 3= 4.2.

The value of f(x) when x=1 is lesser than the value of f(x) when x =0.

So the third graph or bottom left graph represents f(x) = 0.2^{x} + 3.

5 0
3 years ago
Dan and Camille each have a gift card with a combined balance of $350.00. Dan spent 1/2 of his card balance while Camille spent
dezoksy [38]
It should be around $75
3 0
3 years ago
WILL MARK BRAINLIST
iren [92.7K]

Answer:

See below.

Step-by-step explanation:

1/2 = shrink by a factor of 1/2.

+7 = moved up 7 units.

8 0
3 years ago
Quick algebra 1 question for 10 points!
soldier1979 [14.2K]

Using a calculator, the line of best fit for the function is given by:

y = 51.7x - 5.7.

<h3>How to find the equation of linear regression using a calculator?</h3>

To find the equation, we need to insert the points (x,y) in the calculator. For this problem, a linear regression is used because the data only increases.

From the given table, the points are:

(1, 68), (2,97), (3, 134), (4, 176), (5, 241), (6,335).

Inserting these points on the calculator, the line of best fit for the function is given by:

y = 51.7x - 5.7.

More can be learned about a line of best fit at brainly.com/question/22992800

#SPJ1

8 0
2 years ago
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