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3 years ago

To simplify 5.2 + 2.1(4.5 - 41

Mathematics

1 answer:

wlad13 [49]3 years ago

7 0

Answer:

It would be easier to change the decimals to fractions because 41

is a repeating decimal. The decimals should be written as fractions.

Step-by-step explanation:

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Solve for x. Write both solutions, separated by a comma. 5x^2 + 2x -- 7 = 0

tia_tia [17]

Answer: 1 or -7/5

Step-by-step explanation:

5x2+2x-7=0

Factorise

Factors of -35 that will give +2 is +7&-5

5x2+7x-5x-7=0

X(5x+7)-1(5x+7)=0

(X-1)(5x+7)=0

X-1=0

X=1

5x+7=0

5x= -7

X= -7/5

X= 1 or-7/5

4 0

3 years ago

A local bank has determined that the daily balances of the checking accounts of its customers are normally distributed with an a

maria [59]

Answer:

83.43% of customer's balances is between $241 and $301.60.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $280

Standard Deviation, σ = $20

We are given that the distribution of daily balance is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(customer's balances is between $241 and $301.60)

$P(241 \leq x \leq 301.60) \\\\= P(\displaystyle\frac{241 - 280}{20} \leq z \leq \displaystyle\frac{301.60-280}{20}) \\\\= P(-1.95 \leq z \leq 1.08)\\\\= P(z \leq 1.08) - P(z < -1.95)\\\\= 0.8599 -0.0256 = 0.8343 =83.43\%$

Thus, 83.43% of customer's balances is between $241 and $301.60.

5 0

3 years ago

Rationalize the denominator of $\frac{\sqrt{32}}{\sqrt{16}-\sqrt{2}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, whe

musickatia [10]

Rationalizing the denominator involves exploiting the well-known difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

We have

$(\sqrt{16}-\sqrt2)(\sqrt{16}+\sqrt2)=(\sqrt{16})^2-(\sqrt2)^2=16-2=14$

so that

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{32}(\sqrt{16}+\sqrt2)}{14}$

Rewrite 16 and 32 as powers of 2, then simplify:

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{2^5}(\sqrt{2^4}+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{2^2\sqrt2(2^2+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{4\sqrt2(4+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+4(\sqrt2)^2}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+8}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{8\sqrt2+4}7$

So we have A = 8, B = 2, C = 4, and D = 7, and thus A + B + C + D = 21.

3 0

3 years ago

Hello! I need to know which one is equivalent :)

nevsk [136]

A hopefully I'm not mistaking

5 0

4 years ago

Read 2 more answers

Solve for y . 6y=8-9+6y

prisoha [69]

6y=8-9+6y

6y= -1+6y

Move +6y to the left hand .

Sign changes from +6y to -6y

6y-6y= -1+6y-6y

0= -1

Answer: no solution

7 0

3 years ago