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4 years ago

Find the measure of 2

Mathematics

1 answer:

alekssr [168]4 years ago

3 0

Answer:

d 39

Step-by-step explanation:

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What is the GCF of 4x^ 2 +6x

Triss [41]

Answer:

The GCF of 4x^2 + 6x

is. (2x )

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8 0

3 years ago

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F(x)=9x+5, find f^-1(x).

lidiya [134]

Answer:

interchange the role of x and y

x=9y+5

x-5=9y

x-5 =y

so,f^-1(x)= x-5

9

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<h3>stay safe healthy and happy.</h3>

4 0

3 years ago

The width of a triangle is six more than twice the height. The area of the triangle is 88in2. Find the height and base of the tr

malfutka [58]

For this case we have that by definition, the area of a triangle is given by:

$A = \frac {b * h} {2}$

Where:

b: It is the base of the triangle

h: It is the height of the triangle

According to the statement data we have:

$b = 6 + 2h$

Substituting we have:

$88 = \frac {(6 + 2h) * h} {2}\\176 = 6h + 2h ^ 2\\2h ^ 2 + 6h-176 = 0$

We divide between 2 on both sides:

$h ^ 2 + 3h-88 = 0$

We factor by looking for two numbers that, when multiplied, are obtained -88 and when added together, +3 is obtained.

These numbers are +11 and -8.

$(h + 11) (h-8) = 0$

We have two roots:

$h = -11\\h = 8$

We choose the positive value.

Thus, the base of the triangle is: $b = 6 + 2 (8) = 22$

Answer:

The base of the triangle is 22 units.

5 0

4 years ago

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-p(51+z) = Dz+84 Z=?

skad [1K]

Answer:

$z=\frac{3 (28 + 17p)}{p + D}$

Step-by-step explanation:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

8 0

3 years ago

Sin(x2 y2 da r , where r is the region in the first quadrant between the circles with center the origin and radii 1 and 5

Rudik [331]

I assume there's a plus sign missing above...

Convert to polar coordinates, using

$\begin{cases}x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta\end{cases}$

Then the Jacobian is

$\dfrac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}x_r&y_r\\x_\theta&y_\theta\end{vmatrix}=\begin{vmatrix}\cos\theta&\sin\theta\\r\sin\theta&-r\cos\theta\end{vmatrix}=-r$

Then

$\mathrm dA=\mathrm dx\,\mathrm dy=|-r|\,\mathrm dr\,\mathrm d\theta=r\,\mathrm dr\,\mathrm d\theta$

so the integral can be written as

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_1^5r\sin(r^2)\,\mathrm dr\,\mathrm d\theta$

Let $s=r^2$ , so that $\dfrac{\mathrm ds}2=r\,\mathrm dr$ .

$\displaystyle\frac12\int_0^{\pi/2}\int_1^{25}\sin s\,\mathrm ds\,\mathrm d\theta=-\frac12(\cos25-\cos1)\int_0^{\pi/2}\mathrm d\theta=\frac\pi4(\cos1-\cos25)$

3 0

3 years ago