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Ber [7]
3 years ago
10

HELPPP LAST QUESTION!!!!! WILL UPVOTE

Mathematics
1 answer:
RUDIKE [14]3 years ago
6 0
When you multiply the equation 1 by 1, then it won't change and will be:
<span>6x − 2y = 3
Now, when you'll add 2nd equation to it, then, it is:
= 11x + y = 7

In short, Your Answer would be Option D) </span><span>Show that the solution to the system of equations 11x + y = 7 and 5x + 3y = 4 is the same as the solution to the given system of equations

Hope this helps!</span>
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Is anyonee uppppppp or is it just me<br><br> whats 4 + 4 because i dont know
IRISSAK [1]
The answer is 8 lol
3 0
3 years ago
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A customer buys 40 meters of cable. How many centimeters of cable did the customer buy ? (1 meter = 100 centimeters
storchak [24]

Answer:4000

Step-by-step explanation:40 times 100 is 4000

3 0
3 years ago
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jenna has 4 double fudge brownies to share with 5 people including yourself how can jenna share the brownies equally.
Oliga [24]

Answer:

4÷5=0.8      or    4/5

Step-by-step explanation:

6 0
3 years ago
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the graph of F(x), shown below, has the same shape as the graph of G(x)=x^4. but it is shifted 3 units to the right. which is it
Naddik [55]

Answer:

<h2>C. F(x) = (x - 3)⁴</h2>

Step-by-step explanation:

f(x) + n - shift the graph of f(x) n units up

f(x) - n - shift the graph of f(x) n units down

f(x - n) - shift the graph of f(x) n units to the right

f(x + n) - shift the graph of f(x) n units to the left

===================================

The graph og G(x) = x⁴ shifted 3 units to the right.

Therefore F(x) = G(x - 3) = (x - 3)⁴

3 0
3 years ago
A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit
sergiy2304 [10]

Answer:

95% Confidence interval: (39.43, 61.58)

Step-by-step explanation:

We are given the following in the question:

Sample mean, \bar{x} = $50.50

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation = 20

95% Confidence interval:

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.05} = \pm 2.1447  

=50.50 \pm 2.1447(\dfrac{20}{\sqrt{15}} ) \\\\= 50.50 \pm 11.0751 \\= (39.4249,61.5751)\\\approx (39.43, 61.58)  

95% Confidence interval: (39.43, 61.58)

8 0
3 years ago
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