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3 years ago

How do I get 31 to be my answer with placing parentheses somewhere in this expression

Mathematics

1 answer:

pishuonlain [190]3 years ago

8 0

Answer:

4 + 3² × (5 − 2)

Step-by-step explanation:

4 + 3² × (5 − 2)

4 + 3² × 3

4 + 9 × 3

4 + 27

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What is the "middle value" between 10 and 11

Anni [7]

10.5 if I’m not wrong

3 0

3 years ago

Geometry math question

Brut [27]

Look at the picture.

Let |AE| = |AD| = b

We have a proportion:

$\dfrac{x}{13}{y}=\dfrac{19}{y+a}=\dfrac{x}{y+2a}$

Solve for y from first proportion

$\dfrac{13}{y}=\dfrac{19}{y+a}\ \ \ |\text{cross multiply}\\\\19y=13(y+a)\\\\19y=13y+13a\ \ \ \ |-13y\\\\6y=13a\ \ \ |:6\\\\y=\dfrac{13}{6}a$

Substitute to the second proportion

$\dfrac{19}{y+a}=\dfrac{x}{y+2a}\\\downarrow\\\dfrac{19}{\frac{13}{6}a+a}=\dfrac{x}{\frac{13}{6}a+2a}\\\\\dfrac{19}{\frac{13}{6}a+\frac{6}{6}a}=\dfrac{x}{\frac{13}{6}a+\frac{12}{6}a}\\\\\dfrac{19}{\frac{19}{6}a}=\dfrac{x}{\frac{25}{6}a}\ \ \ \ |\dot a\neq0\\\\\dfrac{19}{\frac{19}{6}}=\dfrac{x}{\frac{25}{6}}\\\\19\cdot\dfrac{6}{19}=x\cdot\dfrac{6}{25}\\\\6=\dfrac{6x}{25}\ \ \ |\cdot25\\\\6x=6\cdot25\ \ \ |:6\\\\x=25$

Answer: B. 25

7 0

3 years ago

A student has a savings account earning 3% simple interest. She must pay $1900 for first-semester tuition by September 1 and $19

Anastaziya [24]

Answer:

$3781.19

Step-by-step explanation:

Let us assume that the student has to earn $(1900 + x) by September 1 so that he can pay the $1900 tuition fee by September 1 and the remaining $x will grow at 3% simple interest to make him able to pay another tuition fee of $1900 by January 1.

So, we can write $x( 1 + \frac{4 \times 3}{12 \times 100}) = 1900$

{Because September 1 to January 1 is 4 months and the monthly simple interest rate is $\frac{3}{12}$ %}

⇒ 1.01x = 1900

⇒ x = $1881.19 (Rounded to the nearest cents)

Therefore, the student has to earn $(1900 + 1881.19) = $3781.19 (Answer)

8 0

3 years ago

Integral of x"2+4/x"2+4x+3

dolphi86 [110]

I'm guessing you mean

$\displaystyle \int\frac{x^2+4}{x^2+4x+3}\,\mathrm dx$

First, compute the quotient:

$\displaystyle \frac{x^2+4}{x^2+4x+3} = 1 + \frac{4x-1}{x^2+4x+3}$

Split up the remainder term into partial fractions. Notice that

x ² + 4x + 3 = (x + 3) (x + 1)

Then

$\displaystyle \frac{4x-1}{x^2+4x+3} = \frac a{x+3} + \frac b{x+1} \\\\ \implies 4x - 1 = a(x+1) + b(x+3) = (a+b)x + a+3b \\\\ \implies a+b=4 \text{ and }a+3b = -1 \\\\ \implies a=\frac{13}2\text{ and }b=-\frac52$

So the integral becomes

$\displaystyle \int \left(1 + \frac{13}{2(x+3)} - \frac{5}{2(x+1)}\right) \,\mathrm dx = \boxed{x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C}$

We can simplify the result somewhat:

$\displaystyle x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C \\\\ = x + \frac12 \left(13\ln|x+3| - 5\ln|x+1|\right) + C \\\\ = x + \frac12 \left(\ln\left|(x+3)^{13}\right| - \ln\left|(x+1)^5\right|\right) + C \\\\ = x + \frac12 \ln\left|\frac{(x+3)^{13}}{(x+1)^5}\right| + C \\\\ = \boxed{x + \ln\sqrt{\left|\frac{(x+3)^{13}}{(x+1)^5}\right|} + C}$

3 0

2 years ago

What are the answer 1/7 of 21

HACTEHA [7]

Answer:

The answer is 3.

Step-by-step explanation:

1/7(21)= 3

8 0

2 years ago