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Aneli [31]
2 years ago
12

What is the interval notation for -3<x<4

Mathematics
1 answer:
kvv77 [185]2 years ago
4 0
That would be (-3, 4)
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What is the distance between (-6,4) and (-8,6)? Choose 1 answer:
lukranit [14]

Answer:

D

Step-by-step explanation:

We want to find the distance between (-6, 4) and (-8, 6).

We can use the distance formula given by:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Let (-6, 4) be (x₁, y₁) and let (-8, 6) be (x₂, y₂).

Substitute:

d=\sqrt{(-8-(-6))^2+(6-4)^2

Evaluate:

d=\sqrt{(-2)^2+(2)^2

Evaluate:

d=\sqrt{4+4}=\sqrt{8}

Hence, our answer is D.

6 0
3 years ago
For accounting purposes, the value of assets (land, buildings, equipment) in a business are depreciated at a set rate per year.
lidiya [134]

Answer: $ 125987.80

Step-by-step explanation:

Given: The value, V(t) of $393,000 worth of assets after t years, that depreciate at 15% per year, is given by the formula

V (t)=V_o(b)^t, here V_o is the initial asset value  and b is the multiplicative decay factor.

The exponential decay function is given by ;-

f(x)=A(b)^x, where A is the initial value , x is the  times period and b is the multiplicative decay factor.

where b = 1-r, r is the rate of decay.

Since r = 15%=0.15

Therefore, b = 1-0.15=0.85

Now ,for 7 years , the value of assets is given by :-

V=393000(0.85)^7=125986.795\approx125987.80

Hence, the assets valued at after 7 years = $ 125987.80

4 0
2 years ago
Read 2 more answers
D(n) = -5 (1/2)n-1<br> What is the 3rd term in the sequence?
jonny [76]

Answer:

-5

Step-by-step explanation:

dₙ = -5 (1/2)n-1

the 3rd term in the sequence means n = 3

d₃ = -5(1/2)(3-1) = -5(1/2)(2) = -5

3 0
3 years ago
Read 2 more answers
Hello its me, the 6th grade crack head that needs help on a PEMDAS again! YAYY so anyways the question isssssss 10 divided by (3
Aneli [31]

Answer:

its 8...i think

Step-by-step explanation:

5 0
2 years ago
If y varies directly with x, and y is 12 when x is 8, find y when x is 9.​
jasenka [17]

Answer:

26

Step-by-step explanation:

its not

5 0
2 years ago
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