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Firdavs [7]
2 years ago
14

Could someone Help pleasee ??!

Mathematics
1 answer:
malfutka [58]2 years ago
8 0

Answer:

sorry if im wrong

Step-by-step explanation:

no

no

yes

yes

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Anja represents the inequality 4-g/8 < - 15 with the phrase “the quotient of the difference of 4 and a number and 8 is at lea
scoundrel [369]

Answer:

Anja’s phrase should have ended with “less than –15”, not “at least –15”.

Step-by-step explanation:

Given the statement:

Anja represents the inequality \frac{4-g}{8}< -15

with the phrase “the quotient of the difference of 4 and a number and 8 is at least –15.”

Let the number be g.

“the quotient of the difference of 4 and a number and 8" translated to \frac{4-g}{8}

"at least" means \leq

then;

“the quotient of the difference of 4 and a number and 8 is at least –15.” translated to \frac{4-g}{8}\leq -15

Therefore, Anja’s phrase should have ended with “less than –15”, not “at least –15”.

4 0
3 years ago
Read 2 more answers
(m-3)/(7)=(m)/(m+8) Solve the proportion.
ExtremeBDS [4]

Answer: m=6, m=-4

Step-by-step explanation:

To solve this proportion, we have to cross multiply.

\frac{m-3}{7} =\frac{m}{m+8}

(m-3)(m+8)=7m

Now that we have cross multiplied, we actually need to FOIL the left side to expand the equation.

m^2+8m-3m-24=7m

Combine like terms.

m^2+5m-24=7m

We can move all terms to one side and then solve for m.

m^2-2m-24=0

We can actually factor this to:

(m-6)(m+4)=0

We set each factor equal to 0 to find m.

m-6=0

m=6

m+4=0

m=-4

6 0
3 years ago
Find the average value of the function on the given interval.<br> f(x)=ex/7; [0,7]
Anastaziya [24]

Answer:

f_{avg}=e-1

Step-by-step explanation:

We are given that a function

f(x)=e^{\frac{x}{7}}

We have to find the average value of function on the given interval [0,7]

Average value of function on interval [a,b] is given by

\frac{1}{b-a}\int_{a}^{b}f(x)dx

Using the formula

f_{avg}=\frac{1}{7-0}\int_{0}^{7}e^{\frac{x}{7}} dx

f_{avg}=\frac{1}{7}[e^{\frac{x}{7}}\times 7)]^{7}_{0}

By using the formula

\int e^{ax}=\frac{e^{ax}}{a}

f_{avg}=(e-e^0)=e-1

Because e^0=1

f_{avg}=e-1

Hence, the average value of function on interval [0,7]

f_{avg}=e-1

8 0
3 years ago
Ten less than the cube of a number
Virty [35]

ten less than the cube of a number n:

n^3-10

4 0
3 years ago
WHO EVER CAN ANSWER THESE FOR ME I WILL MARK YO THE BRANLEIST!!!
abruzzese [7]

Answer:

Exercise 4: 15 times

Exercise 5: 5^10

Exercise 9: x^185

Exercise 10: n times

3 0
3 years ago
Read 2 more answers
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