All categories

2 years ago

What is the value of h?

Mathematics

2 answers:

hoa [83]2 years ago

7 0

This is a 45-45-90 special right triangle, the hypotenuse is √2 times the leg. (the two legs are equal.) so the answer is 8√2

Naya [18.7K]2 years ago

5 0

Answer:

1.D 2.C 3.A 4.A 5.C 6.B, Connections Special Right Triangles Practice.

You might be interested in

Let u = <-9, 4>, v = <8, -5>. Find u - v. URGENT

Pie

To subtract vectors, you simply have to subtract correspondent coordinates.

So, if you have

$u = \langle u_1,\ u_2,\ldots u_n\rangle ,\quad v = \langle v_1,\ v_2,\ldots v_n\rangle$

the subtraction is simply

$u-v = \langle u_1-v_1,\ u_2-v_2,\ldots u_n-v_n\rangle$

So, in your case, we have

$\langle-9,4\rangle-\langle 8,-5\rangle=\langle -9-8, 4+5\rangle=\langle -17,9\rangle$

3 0

2 years ago

Help me please cuz some people are senting link when I ask them to not

vladimir2022 [97]

Answer:

the change is 8.4 degrees

Step-by-step explanation:

1.2x7=8.4

3 0

2 years ago

Read 2 more answers

2. Find mzR and mzS. 89° R 157 T Р

baherus [9]

Answer:

m∠R = 123°

m∠S = 91°

Step-by-step explanation:

The measure of an inscribed angle is 1/2 the measure of the intercepted arc

m of arc QRS = 2 m∠P = 2(57) = 114

Now m of arc SPQ = 360 - 114 = 246

So, m∠R = 1/2(246) = 123

m of arc RSP = 2 m∠Q = 2(89) = 178

m of arc PQR = 360 - 178 = 182

m∠S = 1/2(182) = 91

5 0

2 years ago

The safety inspector in a large city wants to estimate the proportion of buildings in the city that are in violation of fire cod

storchak [24]

Answer:

Check Explanation.

Step-by-step explanation:

The sample will be a representative of the entire build us in the city.

For a sample size 35, 3 have fire code violations, hence, the proportion of houses with fire code violations = (3/35) = 0.0857 = 8.57 %

The uncertainty in the estimate is given in the form of standard error.

Standard error = √[(p(1-p)/n]

n = sample size = 35, p = 0.0857, 1 - p = 0.9143

Standard error of the sample = √(0.0857×0.9143/35) = 0.1616

In terms of the population, (0.1616/35) = 0.00462

Proportion of buildings with fire code violations = (8.57 ± 0.462) %

8 0

2 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

2 years ago