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2 years ago

On a piece of paper, graph y = - 2x - 3. Then determine which

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

8 0

Answer:

-3 is where you plot a point then you go up 2 and the the left 1

Step-by-step explanation:

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Can anyone tell me what this is, what unit it is, and how to solve it... Please and thanks. Problem is in the picture.

Vladimir [108]

Answer:

8(+)6=SQR(8^2+6^2)

SQR(64+36)

SQR(100)

=10

Step-by-step explanation:

p + Q is equal to square root of p square + Q Square

sqr means square root

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2 years ago

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1. If Jeremy found 80% of the collector cards and Tim found 41 of the 45 cards,

baherus [9]

Answer:

Tim has more

Step-by-step explanation:

41/45 when you type it into a calculator is 91% and 91%>80%

6 0

2 years ago

What is the y- intercept for the given graph ?

andriy [413]

-1 is the Y intercept for the graph

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2 years ago

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Find the standard form of the equation for the conic section represented by x^2 + 10x + 6y = 47.

Levart [38]

Answer:

The standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Step-by-step explanation:

We know that:

$4p\left(y-k\right)=\left(x-h\right)^2$ is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.

Given the equation

$x^2\:+\:10x\:+\:6y\:=\:47$

Rewriting the equation in the standard form

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Thus,

The vertex (h, k) = (-5, 12)

Please also check the attached graph.

Therefore, the standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

where

vertex (h, k) = (-5, 12)

7 0

2 years ago

Question in pic, will mark brainliest.

bulgar [2K]

Answer:

See explanation

Step-by-step explanation:

Consider triangles ABC and DEC. In these triangles,

$\overline{BC}\cong \overline {EC}$ - given
$\overline{AC}\cong \overline {DC}$ - given;
$\angle ACB\cong \angke DCE$ as vertical angles.

So,

$\triangle ABC\cong \triangle DEC$ by SAS postulate (two sides and angle between these sides of one triangle are congruent to two sides and angle between these sides of another triangle, so triangles are congruent).

Congruent triangles have congruent corresponding parts, hence,

$\overline{BA}\cong \overline{ED}$

3 0

2 years ago