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2 years ago

PLEASE HELP!!!!!!!!!!!!!

Mathematics

1 answer:

mrs_skeptik [129]2 years ago

5 0

If you go from point (-1, -3) to point (4, 4), you can find the slope of the line using the slope formula. $m= \frac{4-(-3)}{4-(-1)}= \frac{7}{5}$ . So our slope is 7/5. The line goes through the y axis at -3/2, so putting that together we get an equation for the line that is $y= \frac{7}{5}x- \frac{3}{2}$ , choice D from above.

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(A) {(-6,2), (3, 4), (4,4), (5, 6)}<br> DOMAIN:<br> RANGE:<br> Is the relation above a function?

vovikov84 [41]

Answer:

Yes

Step-by-step explanation:

Domain: {-6, 3, 4, 5}

Range: {2, 4, 6}

This relation is a function.

6 0

2 years ago

Veronica cut a triangular piece of cookie cake with side lengths that measure 3.6 cm, 6 cm and 4.8cm.

const2013 [10]

Answer:3.6 + 4.8 > 6

Step-by-step explanation:

3 0

3 years ago

Which is bigger 3/8 or 0.3

Luba_88 [7]

$\frac{3}{8} \\
0.3=\frac{3}{10}$

If two fractions have the same numerator, the fraction with the smaller denominator is bigger.
$8\ \textless \ 10 \Rightarrow \frac{3}{8} \ \textgreater \ \frac{3}{10}$

3/8 is bigger.

3 0

2 years ago

Read 2 more answers

Solve. Good luck! Please do not try to google this.

Elena L [17]

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$\frac{x^{2} + 2x - x - 2}{3x(x) - 3x(1)} = \frac{2\sqrt{2x}}{2} + \frac{3x^{2}}{2}$
$\frac{x(x) + x(2) - 1(x) - 1(2)}{3x(x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{x(x + 2) - 1(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$3x(2x - 1)(2\sqrt{2x} + 3x^{2}) = 2(x + 2)(x - 1)$
$3x(2x(2\sqrt{2x} + 3x^{2}) - 1(2\sqrt{2x} + 3x^{2}) = 2(x(x - 1) + 2(x - 1))$
$3x(2x(2\sqrt{2x}) + 2x(3x^{2}) - 1(2\sqrt{2x}) - 1(3x^{2})) = 2(x(x) - x(1) + 2(x) - 2(1)$
$3x(4x\sqrt{2x} + 6x^{3} - 2\sqrt{2x} - 3x^{2}) = 2(x^{2} - x + 2x - 2)$
$3x(4x\sqrt{2x} - 2\sqrt{2x} + 6x^{3} - 3x^{2}) = 2(x^{2} + x - 2)$
$3x(4x\sqrt{2x}) - 3x(2\sqrt{2x}) + 3x(6x^{3}) - 3x(3x^{2}) = 2(x^{2}) + 2(x) - 2(2)$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} + 18x^{4} - 9x^{3} = 2x^{2} + 2x - 4$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} = -18x^{4} + 9x^{3} + 2x^{2} + 2x - 4$
$6x\sqrt{2x}(2x) - 6x\sqrt{2x}(1) = -9x^{3}(2x) - 9x^{3}(-1) + 2(x^{2}) + 2(x) - 2(2)$
$6x\sqrt{2x}(2x - 1) = -9x^{3}(2x - 1) + 2(x^{2} + x - 2)$

5 0

3 years ago

Answer for brainlest for 25 points

Gnom [1K]

Answer:

1. 23/4 , 2. 5.555... , 3. 5 1/2 , 4. √28

you want to make them all into decimals so you can compare then easier

23/4 would be 5.75

√28 (just put it in a calculator) would be 5.29

5 1/2 would be 5.5 (make it into a improper fraction 11/2 then do 11/2 and you get 5.5)

and you have 5.555.

so...

5.75

5.29

5.5

5.55555

first is 5.75 it's bigger because of the .(75)

then 5.555 because it's .(55)

then 5.5 because it's .(50)

and the last one 5.(29)

7 0

2 years ago