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crimeas [40]
3 years ago
14

Solve the linear equation

le="4^{x+7} = 8^{2x-3}" alt="4^{x+7} = 8^{2x-3}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
GREYUIT [131]3 years ago
3 0

Answer:

x = 5.75

Step-by-step explanation:

4^(x+7) = 8^(2x-3)

But; 4^(x+7) = 2^2(x+7)

8^(2x-3) = 2^3(2x-3)

2^2(x+7) = 2^3(2x-3)

Since the bases are the same;

2(x+7) = 3(2x-3)

2x + 14 = 6x -9

14 + 9 = 6x - 2x

23 = 4x

x = 23/4

<u>x = 5.75</u>

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What is the value of z so that -9 and 9 both are solutions of x^2+ z= 103?
siniylev [52]

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Answer: z=22

Explanation:

9^2 + z = 103

81 + z = 103

103 - 81 = z

22 = z

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

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4 0
3 years ago
Simplify 2(3×8-15÷16)<br>​
KengaRu [80]

Answer:

46.125 or 46 1/8

Step-by-step explanation:

2(3×8-15÷16)

=2(24-15÷16)

= 48 - 30/16

= (768 - 30)/16

= 738/16

= 46.125 or 46 1/8

8 0
4 years ago
What should I get at Arby's? Besides the curly fries because I already know those are great
tigry1 [53]

Answer:

bread

Step-by-step explanation:

4 0
3 years ago
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Why is writing rational numbers as a fraction of as decimal useful?
Oxana [17]
Rational Numbers won't be confused into getting to Integers, Whole Numbers, And Counting Numbers. 

Like ALL Counting numbers go into whole numbers, ALL whole numbers go into integers, and ALL integers can go into Rational numbers.

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4 years ago
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prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact
Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable  on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that  limit property of an  integral

\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt

\int_{a}^{b} fdt=f(b)-f(a)....(Equation I)

Using above property of integral then we get

\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt......(Equation II)

Substitute equation I and equation II are equal

Then we get

\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}

\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)

\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)

Therefore, \int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt.

Hence, the value of function does not change after changing the value of function at c.

6 0
3 years ago
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