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3 years ago

Plsss helppp! i will give brainliest!!

Mathematics

2 answers:

geniusboy [140]3 years ago

8 0

It’s the last one hope this guy

yaroslaw [1]3 years ago

6 0

I’m pretty sure it’s the last one (x-7, y+5)

bc if you input (2,3) 2-7, 3+5 it gives you the output -5,8

You might be interested in

Add -4 + (-15). -11 19 -19 11

chubhunter [2.5K]

Step-by-step explanation:

-4+(-15)

=(-19)

I think it helps you

8 0

3 years ago

What is the total surface area??? HELP IM CONFUSED

jeka57 [31]

Answer:

bottom: 12 x 12 = 144

1st triangle: (12 x 8)/2 = 48

2nd triangle: 48

3rd triangle: 48

4th triangle: 48

144 + 48 + 48 + 48 + 48 = 336 cm squared

3 0

3 years ago

I’m having trouble with these types of questions.

german

The sum of all the angles in a triangle should always add up to 360° and seeing that this is an isosceles triangle 2 of the angles should be equal.

So there would be 2 30° angles and 360-60=300 so the 3rd angle would be equal to 300° . Hope this helps!

5 0

3 years ago

Someone, please help me answer this question. I would really appreciate your help. :)

kolezko [41]

Answer:

Step-by-step explanation:

With the hypotenuse being the highest value in a triangle, the other sides must be either the same or lower than the value of the hypotenuse.

4 0

3 years ago

Read 2 more answers

Let F = (2, 3). Find coordinates for three points that are equidistant from F and the y-axis. Write an equation that says P = (x

True [87]

Answer:

The equation that says P is equidistant from F and the y-axis is $P(x,y) =\left(1,\frac{3+y'}{2} \right)$ .

(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.

Step-by-step explanation:

Let $F(x,y) = (2,3)$ and $R(x,y) =(0, y')$ , where $P(x,y)$ is a point that is equidistant from F and the y-axis. The following vectorial expression must be satisfied to get the location of that point:

$F(x,y)-P(x,y) = P(x,y)-R(x,y)$

$2\cdot P(x,y) = F(x,y)+R(x,y)$

$P(x,y) = \frac{1}{2}\cdot F(x,y)+\frac{1}{2} \cdot R(x,y)$ (1)

If we know that $F(x,y) = (2,3)$ and $R(x,y) = (0,y')$ , then the resulting vectorial equation is:

$P(x,y) = \left(1,\frac{3}{2} \right)+\left(0, \frac{y'}{2}\right)$

$P(x,y) =\left(1,\frac{3+y'}{2} \right)$

The equation that says P is equidistant from F and the y-axis is $P(x,y) =\left(1,\frac{3+y'}{2} \right)$ .

If we know that $y_{1}' = -3$ , $y_{2}' = 0$ and $y_{3}' = 3$ , then the coordinates for three points that are equidistant from F and the y-axis:

$P_{1}(x,y) = \left(1,\frac{3+y_{1}'}{2} \right)$

$P_{1}(x,y) = (1,0)$

$P_{2}(x,y) = \left(1,\frac{3+y_{2}'}{2} \right)$

$P_{2}(x,y) = \left(1,\frac{3}{2} \right)$

$P_{3}(x,y) = \left(1,\frac{3+y_{3}'}{2} \right)$

$P_{3}(x,y) = \left(1,6 \right)$

(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.

7 0

3 years ago