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Elan Coil [88]
3 years ago
9

Find quadratic equations with roots -5 and -2

Mathematics
2 answers:
algol133 years ago
8 0

Answer:

y=(x+5)(x+2)

Step-by-step explanation:

Intercept form of quatratics is the best way to solve this equation, since you're given the roots needed.

The intercept form equation is y=a(x-p) (x-q) [where p and q are your roots/intercepts].

So, you input the intercepts and get y=(x-(-5))(x-(-2)), so your final answer should be y=(x+5)(x+2).

Trava [24]3 years ago
7 0

Answer:

y=(x+5)(x+2)

Step-by-step explanation:

Answer:

y=(x+5)(x+2)

Step-by-step explanation:

Intercept form of quadratics is the best way to solve this equation, since you're given the roots needed.

The intercept form equation is y=a(x-p) (x-q) [where p and q are your roots/intercepts].

So, you input the intercepts and get y=(x-(-5))(x-(-2)), so your final answer should be y=(x+5)(x+2).

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Marta_Voda [28]

Answer:

y = 9x + (-7) should be the answer

6 0
3 years ago
An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the radius, in inche
nikklg [1K]
Wow this is a doozy! First you have to figure out what is it you are looking for? If you make a dot in the center of the triangle (which is also the center of the circle) and draw a line from the center to one of the vertices of the triangle you have the radius of the triangle and also of the circle. If you draw all 3 radii from the triangle's center to its vertices, you see you have created 3 triangles within that one triangle.  The trick here is to figure out what your triangle measures are as far as angles go.  If we take the interior measures of those 3 triangles, we get that each one has a measure of 120 (360/3=120). So that's one of your angles, the one across from the side measuring 6. Because of the Isosceles Triangle theorem, we know that the 2 base angles have the same measure because the sides are the same. Subtracting 120 from 180 gives you 60 which, divided in half, makes each of those remaining angles measure 30 degrees. So if we extract that one triangle from the big one, we have a triangle with angles that measure 30-30-120, with the base measuring 6 and each of the other sides measuring 5. If we then split that triangle into 2 right triangles, we have one right triangle with measures 30-60-90. Dropping that altitude to create 2 right triangles not only split the 120 degree angle at the top in half, it also split the base side of 6 in half. So our right triangle has a base of 3 and we are looking for the hypotenuse of that right triangle.  WE have to use right triangle trig for that. Since we have the top angle of 60 and the base of 3, we can use sin60=3/x.  Solving for x we have x=3/sin60 which gives us an x value of 3.5 inches rounded from 3.464. I'm not sure what you mean by a mixed number unless you mean a decimal, but that's the radius of that circle.
6 0
3 years ago
A glass paperweight has a composite shape: a square pyramid fitting exacty on top of an 8 centimeter cube. The pyramid has a hei
ArbitrLikvidat [17]

Answer:

Part 1) The volume of the paperweight is 576\ cm^{3}

Part 2) The total surface area of the paperweight is 400\ cm^{2}

Step-by-step explanation:

Part 1) what is the volume of the paperweight?

we know that

The volume of the paperweight is equal to the volume of the square pyramid plus the volume of the cube

step 1

Find the volume of the pyramid

The volume of the pyramid is equal to

V=\frac{1}{3}BH

where

B is the area of the square base

H is the height of the pyramid

B=8^{2}=64\ cm^{2}

H=3\ cm

substitute

V=\frac{1}{3}(64)(3)=64\ cm^{3}

step 2

Find the volume of the cube

The volume of the cube is equal to

V=b^{3}

V=8^{3}=512\ cm^{3}

step 3

Find the volume of the paperweight

64\ cm^{3}+512\ cm^{3}=576\ cm^{3}

Part 2) what is the total surface area of the paperweight?​

we know that

The total surface area of the paperweight is equal to the surface area of 5 faces of the cube plus the lateral area of the pyramid

step 1

Find the surface area of 5 faces of the cube

SA=5b^{2}

SA=5(8^{2})=320\ cm^{2}

step 2

Find the lateral area of the pyramid

LA=4[\frac{1}{2}bh]

LA=4[\frac{1}{2}(8)(5)]=80\ cm^{2}

step 3

Find the total surface area of the paperweight

320\ cm^{2}+80\ cm^{2}=400\ cm^{2}

8 0
3 years ago
Can someone please help me out ?
gizmo_the_mogwai [7]

Answer:

205-160=<em>d</em>

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
SC.6.E.6.14.
tester [92]
It has to be A I’m sorry if I’m wrong
4 0
2 years ago
Read 2 more answers
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