16 POINTS

1 answer:

8 0

Solve the following system using elimination:
{7 x + 2 y = -19 | (equation 1)
{2 y - x = 21 | (equation 2)

Add 1/7 × (equation 1) to equation 2:
{7 x + 2 y = -19 | (equation 1)
{0 x+(16 y)/7 = 128/7 | (equation 2)

Multiply equation 2 by 7/16:
{7 x + 2 y = -19 | (equation 1)
{0 x+y = 8 | (equation 2)

Subtract 2 × (equation 2) from equation 1:
{7 x+0 y = -35 | (equation 1)
{0 x+y = 8 | (equation 2)

Divide equation 1 by 7:
{x+0 y = -5 | (equation 1)
{0 x+y = 8 | (equation 2)

Collect results:
Answer: {x = -5, y = 8

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Lisa made a shirt using 1/3 m of blue fabric and 3/5 m of red fabric how many meters of fabric did she use in all

Marysya12 [62]

Answer: 14/15 of a meter

Step-by-step explanation:

5 and 3 LCM is 15.

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5 0

2 years ago

The number of decimal places in the product will be sum of the factor"s decimal place.So the product will have 1,2,3 or 4 decima

vampirchik [111]

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$0.47 \to 2\ digits$

$3.2 \to 1\ digit$

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Step-by-step explanation:

Given

$0.47 * 3.2 =$

See comment for complete question

Required

The number of decimal places

To do this, we simply calculate the number of digits after the decimal points

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$3.2 \to 1\ digit$

The product is the sum of the digits above 2 + 1 = 3

Hence:

$Product \to 3\ digits$

6 0

3 years ago

suppose that you have 3000$ to invest. which investment yields the greater return over a 10 year period: 8.04% compounded daily

Ksenya-84 [330]

Answer: Option A: 8.04% compounded daily

<u>Step-by-step explanation:</u>

$A = P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\qquad where\\\\\bullet A = \text{accrued amount (principal plus interest earned)}\\\bullet P = \text{principal (amount invested)}\\\bullet r = \text{rate (in decimal form)}\\\bullet n=\text{number of times compounded in one year}\\\bullet t=\text{time (number of years)}\\\\\\Option\ A:\\A = \text{unknown}\\P=3000\\r=8.04\%\rightarrow 0.0804\\n=\text{daily}\rightarrow 365\\t=10\\\\A=3000\bigg(1+\dfrac{0.0804}{365}\bigg)^{365\times 10}\\\\.\ =\$ 6,702.79$

$Option\ B:\\A = \text{unknown}\\P=3000\\r=8.1\%\rightarrow 0.081\\n=\text{quarterly}\rightarrow 4\\t=10\\\\A=3000\bigg(1+\dfrac{0.081}{4}\bigg)^{4\times 10}\\\\.\ =\$ 6,689.37$