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3 years ago

Can somebody help me with this problem!? i need help ahh

Mathematics

2 answers:

olya-2409 [2.1K]3 years ago

7 0

Answer:

134 Degrees.

Step-by-step explanation:

Lets figure out the third angle (y) in the triangle first.

70 + 64 + y = 180

134 + y = 180

y = 46

We now know that 46 is our third angle. Therefore, we can subtract 46 (y) from 180 (the total between x and the third angle) to get our final answer. 134 Degrees.

slamgirl [31]3 years ago

6 0

for these kinds of problems the outside angle always equals the opposite 2. meaning angle x=64+70 which is 134 degrees. you can double check this because all triangles equal 180 degrees. meaning that the last angle in the triangle is 46. 46+70+64=180.

so your answer is 134.

hope this helps :)

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Pls help me!!!!!!!!!

Aliun [14]

Answer:

The perimeter is 30 units

Step-by-step explanation:

To find the perimeter of SOW, what we need to do is to add all the lengths around the triangle together.

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3 years ago

[Help asap] are these congruent?

PolarNik [594]

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3 years ago

Read 2 more answers

In the diagram, what is the measure of angle 1?

VLD [36.1K]

We know that
angle (3x) and angle (9x) are supplementary angles
so
3x+9x=180°------> 12x=180°------> x=180°/12-----> x=15°

angle (9x) and angle (1) are supplementary angles
so
9x+∡1=180---------> 9*15+∡1=180
∡1=180-9*15---------> ∡1=180-135------> ∡1=45°

the answer is
∡1 is 45°

alternative method
angle 1 = angle 3x----------> vertical angles
∡1=3x-----> 3*15-----> 45°

7 0

3 years ago

I need help with this question! (it's the picture)

Sauron [17]

Cos(38°) = x/44 ⇒ x = 44 * cos(38°) ≈ 34.7 units

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3 years ago

Please help with this calculus problem!

ELEN [110]

$\dfrac{\mathrm dQ}{\mathrm dt}=\dfrac{a(1-5bt^2)}{(1+bt^2)^4}$
$Q(t)=\displaystyle\int\frac{a(1-5bt^2)}{(1+bt^2)^4}\,\mathrm dt$

Let $t=\dfrac1{\sqrt b}\tan u$ , so that $\mathrm dt=\dfrac1{\sqrt b}\sec^2u\,\mathrm du$ . Then the integral becomes

$\displaystyle\frac a{\sqrt b}\int\frac{1-5b\left(\frac1{\sqrt b}\tan u\right)^2}{\left(1+b\left(\frac1{\sqrtb}\tan u\right)^2\right)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{(1+\tan^2u)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{\sec^6u}\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(\cos^6u-5\sin^2u\cos^4u)\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(6\cos^2u-5)\cos^4u\,\mathrm du$

One way to proceed from here is to use the power reduction formula for cosine:

$\cos^{2n}x=\left(\dfrac{1+\cos2x}2\right)^n$

You'll end up with

$=\displaystyle\frac a{16\sqrt b}\int(5\cos2u+8\cos4u+3\cos6u)\,\mathrm du$
$=\displaystyle\frac a{16\sqrt b}\left(\frac52\sin2u+2\sin4u+\frac12\sin6u\right)+C$
$=\dfrac a{32\sqrt b}(5\sin2u+4\sin4u+\sin6u)+C$
$Q(t)=\dfrac{at}{(1+bt^2)^3}+C$

6 0

3 years ago