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Pachacha [2.7K]
2 years ago
10

[(8*9)] - (6*7)] -15

Mathematics
2 answers:
strojnjashka [21]2 years ago
8 0

[(8*9)] - (6*7)] -15

do the parenthesis first because of using PEMDAS

P= parenthesis

E= exponents

M= multiplication

A= addition

S= subtraction

(8*9)=72

(6*7)=42

72-42=30

72-42-15

30-15

=15

Answer: 15

maks197457 [2]2 years ago
3 0

Answer:

15

Step-by-step explanation:

Simplify the following:

8×9 - 6×7 - 15

8×9 = 72:

72 - 6×7 - 15

-6×7 = -42:

72 + -42 - 15

72 - 42 - 15 = 72 - (42 + 15):

72 - (42 + 15)

| 4 | 2

+ | 1 | 5

| 5 | 7:

72 - 57

| 6 | 12

| 7 | 2

- | 5 | 7

| 1 | 5:

Answer:  15

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Step-by-step explanation:

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If the two equations in parenthesizes were there own equations which ones would equal 1 and 6
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3 years ago
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Evaluate the function at the given numbers( correct to six decimal places). Use the results to guess the value of the limit or e
netineya [11]

Answer:

Value of the limit is 0.5.

Step-by-step explanation:

Given,

F(x)=\frac{e^x-1-x}{x^2}

When,

x=1,F(1)=frac{e^1-1-1}{1}=e-2=0.718281

x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885

x=0.1, F(0.1)=\frac{e^0.1-1-0.1}{(0.1)^2}=0.517091

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x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)

Correct upto six decimal places.

Now,

\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}   (\frac{0}{0}) form, applying L-Hospital rule that is differentiating numerator and denominator we get,

\lim_{x\to 0}F(x)

=\lim_{x\to 0}\frac{e^x-1}{2x}    (\frac{0}{0}) form.

=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)

Limit exist and is 0.5. That is according to (1) we can see as the value of x lesser than 1 and tending to near 0, value of the function decreases respectively. And from (2) it shows ultimately it decreases and reach at 0.5, consider as limit point of F(x).  

8 0
3 years ago
I need help with these three problems please​
Liono4ka [1.6K]

Answer:

Q7. 11.3 inches (3 s.f.)

Q8. 96.2 ft

Q9. 36.4cm

Step-by-step explanation:

Q7. Please see attached picture for full solution.

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Applying Pythagoras' Theorem,

34^{2}  =  {x}^{2}  +  {x}^{2}  \\ 2 {x}^{2}  = 1156 \\  {x}^{2}  = 1156 \div 2 \\  {x}^{2}  = 578 \\ x =  \sqrt{578}  \\

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Since the perimeter of the equilateral triangle is 126cm,

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The green line drawn in picture 3 is the altitude of the triangle.

Let the altitude of the triangle be x cm.

sin 60°= \frac{x}{42}

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(to 3 s.f.)

Therefore, the length of the altitude of the triangle is 36.4cm.

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