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2 years ago

Subtract -46-(-19) the answer

Mathematics

1 answer:

tatuchka [14]2 years ago

6 0

Just apply sign operations, remember:
negative times negative gives positive
-46 - (-19) = -46 + 19 = -27

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The shapes below are mathematically similar.

Lera25 [3.4K]

Answer:

Triangle small ABC

Triangle big DEF

AB = 8 cm , BC = 12 cm , DE = 20 cm , EF = x

So;

$\frac{AB}{DE} = \frac{BC}{EF} \\ \\ \frac{8}{20} = \frac{12}{x} \\ \\ 8x = 12 \times 20 \\ 8x = 240 \\ \\ x = \frac{240}{8} \\ \\ x = 30 cm$

I hope I helped you^_^

7 0

2 years ago

Verify which of the following are identities.

-Dominant- [34]

Answer:

Only the second equation is an identity

Step-by-step explanation:

$8 \frac{\tan^2(\theta)}{\sec(\theta)} \csc^2(\theta)=8 \csc(\theta)$

$\tan^2(\theta)\csc^2(\theta)=\sec^2(\theta)$

<u>You can confirm it: </u>

$\frac{\sin^2(\theta)}{\cos^2(\theta)}\cdot \frac{1}{\sin^2(\theta)}= \frac{1}{cos^2(\theta)}= \sec^2(\theta)$

<u>Therefore</u>

$8 \frac{\sec^2(\theta)}{\sec(\theta)} =8 \csc(\theta)$

$8 \frac{\sec(\theta)}{1} =8 \csc(\theta)$

$8\sec(\theta)=8\csc(\theta)$

<h2>It is not an Identity</h2>

<u>Let's the second one</u>

$13 \frac{\tan^2(\theta)}{\sec(\theta)} \csc^2(\theta)=13\sec(\theta)$

In this case, we already performed the calculations, so it is true. It is an Identity.

3 0

2 years ago

50 POINTS!

Hoochie [10]

Answer:

Solution given:

AB=3units

<B=45°

<C=63°

DF=3units

<E=63°

<F=45°

for congruence opposite corresponding angle or side should be equal

∆ABCis congruent to ∆DFE by A.A.S axiom

Similarly

its another name also be

∆BAC=~ ∆FDE

you got it.third one.

7 0

2 years ago

Read 2 more answers

What is the length of PR?

kicyunya [14]

Answer:

$8\sqrt{3}$

Step-by-step explanation:

2/3 pi radians = 2/3 * 180 = 2 * 60 = 120 degrees

split triangle in half:

30, 60, 90 triangle

1/2 PR= 4 root 3

7 0

2 years ago

Two college roommates have each committed to donating to charity each week for the next year. The roommates’ weekly incomes are

schepotkina [342]

Answer:

C. 5 weeks.

Step-by-step explanation:

In this question we have a random variable that is equal to the sum of two normal-distributed random variables.

If we have two random variables X and Y, both normally distributed, the sum will have this properties:

$S=X+Y\\\\\ \mu_S=\mu_X+\mu_Y=30+60=90\\\\\sigma_S=\sqrt{\sigma_X^2+\sigma_Y^2}=\sqrt{10^2+20^2}=\sqrt{100+400}=\sqrt{500}=22.36$

To calculate the expected weeks that the donation exceeds $120, first we can calculate the probability of S>120:

$z=\frac{S-\mu_S}{\sigma_S} =\frac{120-90}{22.36}=\frac{30}{22.36}=1.34\\\\P(S>120)=P(z>1.34)=0.09012$

The expected weeks can be calculated as the product of the number of weeks in the year (52) and this probability:

$E=\#weeks*P(S>120)=52*0.09012=4.68$

The nearest answer is C. 5 weeks.

6 0

2 years ago