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mote1985 [20]
3 years ago
8

What type of angle is formed by the base and height of a triangle?​

Mathematics
1 answer:
Harrizon [31]3 years ago
3 0

Answer:

the Angle is an Acute

Step-by-step explanation:you can't but a little box in it and it it not a 90 degree angle

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suppose a parabola has a vertex of (-4 7) and also passes through the point (-3,8). what is the equation of the parabola in vert
jasenka [17]
The equation would be y = (x+4)² + 7

Because adding 4 to 'x' creates a transformation 4 units left and adding 7 to the whole function moves the parabola up 7

Hope this helps!
8 0
3 years ago
Read 2 more answers
What are these questions i need help
Over [174]

The roots of the equation f(x) = x^2 - 93987 are x = √93987 and x = -√93987

<h3>What are quadratic equations?</h3>

Quadratic equations are equations that have a second degree and have the standard form of ax^2 + bx + c = 0, where a, b and c are constants and the variable a does not equal 0

<h3>How to determine the other roots of the equation?</h3>

The equation of the function is given as:

f(x) = x^2 - 93987

The above equation is a quadratic equation

Express the equation as a difference of two squares

f(x) = (x - √93987)(x + √93987)

Set the equation of the function to 0

(x - √93987)(x + √93987) = 0

Split the factors of the above function equation as follows

x - √93987 = 0 and x + √93987 = 0

Solve for x in the above equations

x = √93987 and x = -√93987

Hence, the roots of the equation f(x) = x^2 - 93987 are x = √93987 and x = -√93987

Read more about roots of equation at

brainly.com/question/776122

#SPJ1

5 0
1 year ago
Please help me 17+2(7+3)-2^2
drek231 [11]

Answer:

t&hhhheeeee awwwwwnnnnssseeerrr is 33

3 0
3 years ago
What is the value of x? enter your answer in the box.
Mamont248 [21]

Answer:

if 9x8 =72

then 56/8= 7

x=7



8 0
3 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
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