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3 years ago

An amount obtained as a result of subtracting numbers

Mathematics

1 answer:

SpyIntel [72]3 years ago

4 0

Answer:

difference

An amount obtained as a result of subtracting numbers is known as difference

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What is the median of the ages in this stem-and-leaf plot? 43 45 46 47

ICE Princess25 [194]

To find a median of an even amount of numbers, add the two middle numbers, and divide by two.

45 + 46 = 91

91/2 = 45.5

45.5 is your median

hope this helps

6 0

3 years ago

Read 2 more answers

Help plz:)))I’ll mark u Brainliest !

seropon [69]

Answer:

A= 18.4

Step-by-step explanation:

i dont think you'll need to show work so ill save u time :)

4 0

3 years ago

Read 2 more answers

CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION

Murljashka [212]

Answer:

The value of DC is 66.34.

Step-by-step explanation:

The triangles ABC and DCA are right angled triangles.

The straight line AC is a bisector for angles C and A.

The measure of ∠C is 30°.

Then the measure of angles BCA and ACD will be 15° each.

The measure of angle DAB is 150°.

Then the measure of angles DAC and BAC will be 75° each.

Now consider the right angled triangle ABC.

The measure of side AC is:

$AC^{2}=AB^{2}+BC^{2}\\AC=\sqrt{AB^{2}+BC^{2}}\\=\sqrt{4^{2}+(10\sqrt{3})^{2}}\\=\sqrt{316}$

Consider the right angled triangle DCA.

The angle DAC measure 75°.

Using the trigonometric identities compute the value of Perpendicular DC as follows:

$tan\ 75^{o}=\frac{DC}{\sqrt{316}}\\\\DC=3.732\times\sqrt{316}\\\\DC=66.34$

Thus, the value of DC is 66.34.

7 0

3 years ago

What does 2n − 3x equally to

DIA [1.3K]

Answer:

2÷45 - 76 and the answer is yours

5 0

2 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago