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3 years ago

Sides and angles of this triangle?

Mathematics

2 answers:

ANEK [815]3 years ago

8 0

The sides are AB, BC, and AC. AC is about 1.48 in length. The angles are A B and C and are unknown other than angle B is 90 degrees.

mestny [16]3 years ago

5 0

A to B is 5, A to C is 9, and B to C is 3

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Bidmas 7+8 divided 5-2

IrinaVladis [17]

Answer:

Step-by-step explanation:

7 + 8 = 15

5 - 2 = 3

∴ 7 + 8 ÷ 5 - 2

= 15 / 3

= 5

Hope this helps

plz mark as brainliest!!!!!!

6 0

3 years ago

Read 2 more answers

On a field trip, there are 3 adults for every 45 students. Which graph models a relationship

Georgia [21]

Answer:

H (the second graph)

Step-by-step explanation:

5 0

3 years ago

A 20-gallon container is filled halfway with a mixture that is $90\%$ vinegar and $10\%$ water. How many gallons of water must b

Yuki888 [10]

Answer:

5 gallons of water must be added.

10 gallons to start with

adding 'x' gallons of water final volume = 10 + x

.9 (10) = amount of vinegar

.1(10) = amount of water

amount of vinegar before and after water addition is constant

.9(10) = .6(10+x)

9 = 6 + .6x

3 = .6x

x = 5 gallons of water to be added

(i truly hope i helped you

8 0

2 years ago

Help please/////////////////////

Wewaii [24]

i dont understand the question

7 0

2 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

2 years ago