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3 years ago

The graph shows quadrilaterals q and r

Mathematics

2 answers:

tatuchka [14]3 years ago

8 0

Answer:thanks

Step-by-step explanation:

irina1246 [14]3 years ago

4 0

Answer:

Djudhdxhd

Step-by-step explanation:

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Insert a monomial so that each result is an identity( *− 3b4)(3b4 +*) = 121a10 − 9b8

natita [175]

The answer for this is (-3b4) (3b4 +) = 121a10-9b8

6 0

2 years ago

A triangle has side lengths of 23 in, 19 in, and 37 in

ArbitrLikvidat [17]

Formula is bh/2.
So, 23*19/2 = 218.5

4 0

3 years ago

What is the area of a 45 degree sector of a circle with a radius of 12 in.

Elodia [21]

given ,

a circle of radius 12 inches

and $\theta$ = 45°

now we know that ,

$\\{Area \: of \: sector = \frac{\theta}{360\degree} \times \pi \: r {}^{2} } \\ \\$

let's now plug in the values of radius and theta as 12 inches and 45° respectively ,

$\\\dashrightarrow \: \frac{45}{360} \times \frac{22}{7} \times 12 \times 12 \\ \\ \dashrightarrow \: \frac{1}{8} \times \frac{22 \times 12 \times 12}{7} \\ \\ \dashrightarrow \: \frac{22 \times 12 \times 12}{56} \\ \\ \dashrightarrow \: \frac{3168}{56} \\ \\ \dashrightarrow \: 56.57 \: inches {}^{2} (approx.)$

hope helpful :D

6 0

2 years ago

Find the point on the graph of y=x^2 where the tangent line has slope m= - 1/3

babunello [35]

Here the graph y=x²

Taking the 1st derivative here of the y it will be

y1 = 2x

m=-1/3

That is m=2x

so x=-1/6

so y=(-1/6)²=1/36

So the coordinate of point will be=(-1/6 , 1/36)

8 0

3 years ago

Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, -5). Write the equati

amid [387]

In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
$f(x) = a(x - x_1)(x-x_2)(x-x_3)$

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form $(x_n , 0)$

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say $x_1 = 2 , x_2 = 3 , x_3 = 5$

So required equation is
$f(x) = a (x-2)(x-3)(x-5)$
              $= a[(x^2 - 2x - 3x + 6)(x-5)]$
              $= a[(x^2 - 5x+6)(x-5)]$
              $= a(x^3 - 5x^2 + 6x- 5x^2 + 25x - 30)$
              $= a(x^3 - 10x^2+31x-30)$
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
$-5= a (0-0+0 - 30)$
$-5 = -30a$
$a = \frac{-5}{-30} = \frac{1}{6}$
So required equation of cubic polynomial is
$f(x) = \frac{1}{6}(x^3 -10x^2+31x-30)$

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.

4 0

3 years ago

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