All categories

3 years ago

What is the answer??

Mathematics

1 answer:

Svetradugi [14.3K]3 years ago

6 0

Answer:

32.

Step-by-step explanation:

I accidentally put my age too high. Im only in middle school but 32 could be one answer. If two sides are the same then it could be an isoceles triangle or whatever.

You might be interested in

PLEASE HELP ME I NEED THE PRIME FACTORIZATION!!!!!!!!!!!!!!!

docker41 [41]

Hey there :)

To find the Greatest Common Factor ( GCF ) of 36, 40 and 45, let us first separate:
Factors of 36: 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
Factors of 40: 1 , 2 , 4 , 5 , 8 , 10 , 20 , 40
Factors of 45: 1 , 3 , 5 , 9 , 15 , 45

Therefore, the GCF is 1 since only that is common in all three.

To find the Least Common Multiple of 36, 40 and 45, we see which multiplication matches in the least:
36 × 1 = 36 | 40 × 1 = 40 | 45 × 1 = 45
36 × 2 = 72 | 40 × 2 = 80 | 45 × 2 = 90
36 × 3 = 108 | 40 × 3 = 120 | 45 × 3 = 135
36 × 4 = 144 | 40 × 4 = 160 | 45 × 4 = 180
36 × 5 = 180 | 40 × 5 = 200 | 45 × 5 = 225
36 × 6 = 216 | 40 × 6 = 240 | 45 × 6 = 270
36 × 7 = 252 | 40 × 7 = 280 | 45 × 7 = 315
36 × 8 = 288 | 40 × 8 = 320 | 45 × 8 = 360
36 × 9 = 324 | 40 × 9 = 360
36 × 10 = 360

Therefore, the LCM is 360

4 0

3 years ago

The velocity of an object is given by v(t)=t(t^2-1)^(1/3) feet per second. Find the total number of feet traveled by the object

NARA [144]

What would u like help with

7 0

4 years ago

Solve Using Dirac Deltla/discontinuous forcing

Jet001 [13]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're told that $A(0)=0$ .

For $0\le t\le15$ , the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for $t>15$ . We can capture this in terms of the unit step function $u(t)$ and Dirac delta function $\delta(t)$ as

$\text{rate in}=u(t)-u(t-15)+20\delta(t-15)$

(in lb/min)

The salt from the mixed solution flows out at a rate of

$\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}$

Then the amount of salt in the tank at time $t$ changes according to

$\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}$

Let $\hat A(s)$ denote the Laplace transform of $A(t)$ , $\hat A(s)=\mathcal L_s\{A(t)\}$ . Take the transform of both sides to get

$s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)$

Solve for $\hat A(s)$ , then take the inverse of both sides.

$\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}$

$\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}$

7 0

4 years ago

BRAINLIEST AND TEN POINTS TO CORRECT ANSWER AND NO (5,2) WAS WRONG PLEASE HELP !!

lana66690 [7]

Answer: 5,4

Step-by-step explanation:

I did the test

7 0

3 years ago

M5_A3. The expected project completion time for the construction of a pressure vessel is 18 monthsand the project variance is 7.

snow_tiger [21]

Answer:

a) 0.3520

b) 0.9884

c) 0.1292

d) 0.1292

e) 0.7648

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

The expected project completion time for the construction of a pressure vessel is 18 months. So $\mu = 18$ .

And the project variance is 7. The standard deviation is the square root of the variance, so $\sigma = \sqrt{7}$ .

a) What is the probability that the project will require at least 19 months?

The pvalue of the zscore of $X = 19$ is the probability that the project will require less than 19 months. So the probability that the project will require at least 19 months is the subtraction of 1 by this pvalue.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{19 - 18}{\sqrt{7}}$

$Z = 0.38$

The pvalue of Z = 0.38 is 0.6480.

So the probability that the project will require at least 19 months is 1 - 0.6480 = 0.3520.

b) What is the probability that the project will be completed within 24 months?

This is the pvalue of the zscore of X = 24.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24 - 18}{\sqrt{7}}$

$Z = 2.27$

The pvalue of Z = 2.27 is .9884. This means that the probability that the project will be completed within 24 months is .9884.

c) What is the probability that the project will require at least 21 months?

Same logic as a). We find the pvalue of the zscore of X = 21, then we subtract 1 by this value.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21 - 18}{\sqrt{7}}$

$Z = 1.13$

The pvalue of Z = 1.13 is 0.8708.

So the probability that the project will require at least 21 months is 1 - 0.8708 = 0.1292.

d) What is the probability that the project will be completed within 15 months?

Same logic as b). This is the pvalue of the zscore of X = 15.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{15 - 18}{\sqrt{7}}$

$Z = -1.13$

The pvalue of Z = -1.13 is .1292. This means that the probability that the project will be completed within 15 months is .1292.

e) What is the probability that the project will take between 16 and 24 months to complete?

This is the pvalue of Z = 24 subtract by the pvalue of Z = 16.

In b), we found that the pvalue of Z = 24 is .9884

Now we find the pvalue of Z = 16

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16 - 18}{\sqrt{7}}$

$Z = -0.76$

The pvalue of Z = -0.76 is 0.2236

So the probability that the project will take between 16 and 24 months to complete is 0.9884 - 0.2236 = 0.7648

8 0

4 years ago