All categories

3 years ago

Hurrry i need help nowwwwwwwwwwwwwwwww

Mathematics

2 answers:

polet [3.4K]3 years ago

4 0

The answer is A
hope this helps ;)

nikklg [1K]3 years ago

3 0

Answer: A y=4x^2+1

Step-by-step explanation:According to slope A is the answer

You might be interested in

Which equation is true about the measure of the angles of the triangle?the measure of angle prq is equal to 100 degreesthe measu

harkovskaia [24]

I believe this question is missing an image. However, I will just tell how how to solve such question and you can apply this solution to the triangle you have.

In any given triangle, the sum of the measure of the three angles must be equal to 180 degrees.
Therefore, for a triangle pqr:
measure angle pqr + measure angle prq + measure angle rqp = 180 degrees

If you have one angle given, then the sum of the other two angles will be equal to 180 - measure of the given angle.

If you have two angles given, then the measure of the third angle will be equal to 180 - the sum of the other two angles.

3 0

2 years ago

After working for a few years, the instructor has $3,500 saved for rent. He rents an apartment that costs $275 a month. Which gr

djverab [1.8K]

Answer: c if you know where i mean

8 0

2 years ago

Read 2 more answers

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

2 years ago

I need help can you guys help me out thank you!! p.s. the word is meaters

Gala2k [10]

It’s 76 meters hoped it helped!

6 0

2 years ago

What is the answer 47 * 63

Anuta_ua [19.1K]

<h3>U can also verify the answer using a calculator </h3>

8 0

2 years ago