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3 years ago

(PLEASE HELP AS SOON AS POSSIBLE)

Mathematics

2 answers:

stiv31 [10]3 years ago

6 0

opt G. (-4,-2)

I don't know...

Mkey [24]3 years ago

3 0

Answer:

Step-by-step explanation:

The origin in a graph is the point (0, 0) where x and y are both 0. If you move 4 units left, x is -4. And if you go down 2 units, y is -2. So the ordered pair would be (-4, -2)

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Did I get this right or no I didn’t

Nezavi [6.7K]

Answer:

its correct bro ;) click next

7 0

3 years ago

Dianna and Becky were playing on the same soccer team and took turns being goalie. They stoped 9 out of 10 shots made against th

vitfil [10]

Answer:

27 balls

Step-by-step explanation:

If Dianna and Becky stopped 9 out of 10 shots made against them.

It means they allowed in 1 out of 10 shots made against them.

This is best solved using ratios

Proportion of Shots Stopped : Proportion of Shots allowed=9:1

If the other team scores 3 points, it means the number of shots allowed =3

Let x be the number of shots stopped

Number of Shots Stopped : Number of Shots allowed=x:3

Therefore:

9:1=x:3

$\frac{9}{1}=\frac{x}{3}$

Cross multiplying

x=9 X 3 =27

Dianna and Becky stopped 27 balls from going into the net,

6 0

3 years ago

Has a slope of -2/5 and goes through (-10,-1)

Aleonysh [2.5K]

Answer:

Below in bold.

Step-by-step explanation:

Using the point-slope form of a straight line equation:

y - y1 = m(x - x1)

y - (-1)) = -2/5(x - -10)

y + 1 = -2/5(x + 10)

y + 1 = -2/5x - 4

y = -2/5x - 5.

In standard form this is:

2x + 5y = -25.

7 0

2 years ago

A spinner is separated into 4 equal pieces, as shown below:

kakasveta [241]

Getting black is 1/4 of the percentage, and spinning it 10 times gives us 10/40, because there are 40 possible colors to land on when u spin it 10 times. The last spin, is just like any other time you spin it. just because it's the last spin, doesn't make the black sector greater. So the answer is 1/4 or 25%.

8 0

3 years ago

X^3+7x^2+13x+4=0 Show all work

Softa [21]

$\displaystyle\\
x^3+7x^2+13x+4=0\\\\
x^3+\underbrace{4x^2+3x^2}_{7x^2}+\underbrace{12x+x}_{13x}+4=0\\\\
(x^3+4x^2) + (3x^2 + 12x)+(x+4)=0\\\\
x^2(x+4)+3x(x+4)+(x+4)=0\\\\
(x+4)(x^2 + 3x +1)=0\\\\
x+4 = 0~~~\Longrightarrow~~~\boxed{x_1 = -4}\\\\
x^2 + 3x +1=0\\\\
x_{23}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}= \frac{-3\pm \sqrt{3^2-4\cdot 1 \cdot 1} }{2\cdot 1}= \frac{-3\pm \sqrt{5} }{2}\\\\
\boxed{x_2 = \frac{-3-\sqrt{5} }{2}}\\\\
\boxed{x_3 = \frac{-3+\sqrt{5} }{2}}$

4 0

3 years ago