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JulsSmile [24]
2 years ago
11

Tell me the Definition of each Word listed (math)

Mathematics
1 answer:
natita [175]2 years ago
3 0
Constant: A value that doesn't change. Instead, it's a fixed value.

Variable: A symbol (usually a letter) standing in for an unknown numerical value in an equation.

Term: Either a single number or variable, or numbers and variables multiplied together. (Terms are separated by + or − signs, or sometimes by divide.)

Like Terms: Terms whose variables (and their exponents such as the 2 in x2) are the same. In other words, terms that are "like" each other.

Coefficient: A number used to multiply a variable.
You might be interested in
Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places
natka813 [3]

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

\rm 5 \sin \theta  = 4\\\\\theta = 0.927 , 2.214

Then by the integration, we have

\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\

\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\

On solving, we have

\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

#SPJ4

5 0
2 years ago
One of your friends lives 225 km away. It took him 2.5 hours to travel to the cottage. At what speed did he travel?
Dahasolnce [82]
Speed = distance / time

given:
distance = 225 km and time = 2.5 hours

so
speed = 225 / 2.5 = 90 km per hour

Answer:
Speed = 90 km per hour
8 0
2 years ago
= 4. The sum of Ben's and David's ages is 62 yrs now. Five years from now David will be twice as old as Ben. How old is Ben now?
ki77a [65]

The age of ben and david is 24 and 38 years.

According to the statement

we have to find the age of the Ben now.

So,For this purpose, we know that the

One application of linear equations is what are termed age word problems. When solving age problems, generally the age of two different people (or objects) both now and in the future (or past) are compared.

From the given information:

The sum of Ben's and David's ages is 62 yrs now

And

Five years from now David will be twice as old as Ben.

Let the ben age is x

and the david age is y.

So, from given information:

x+y = 62  -(1)

5 years from now,

y = 2x  -(2)

put (2) in the (1) then

x+y = 62 -5

x+2x = 57

3x = 57

x = 19.

And the age of davud become

y = 2(19)

y = 38 years

So, five years from now the age of ben is 19 and david is 38 years.

And now the age of the ben and age of david is

Age of ben now = 19+5

Age of ben now = 24.

Age of david now  = 62 -24

Age of david = 38.

So, The age of ben and david is 24 and 38 years.

Learn more about age word problems here

brainly.com/question/13818690

#SPJ9

5 0
1 year ago
83 tens times 2 tens
gizmo_the_mogwai [7]

Answer:

16600 should be the answer.

Step-by-step explanation:

because 830 times 20 is 16600.

hope i helped

6 0
3 years ago
What is the other square root of 119 + 120i?
MakcuM [25]

Answer:

\sqrt{119+120 i}=\pm (12.24+i 4.90)

Step-by-step explanation:

Given

z = 119 + 120 i

Let \sqrt{119+120 i}=p+iq

Squaring both sides

119+120 i=p^2-q^2+2ipq

Comparing real and imaginary part

Re(LHS)=Re(RHS)

119=p^2-q^2...........................(1)

comparing Im(LHS)=Im(RHS)

120=2pq

q=\frac{60}{p}

Substitute q in 1

119=p^2-(\frac{60}{p})^2

p^4-119p^2-(68)^2=0

Let x=p^2

x^2-119x-4624=0

x=\dfrac{119\pm \sqrt{119^2+4\times 4624}}{2}

x=\frac{119\pm 180.71}{2}

we take only Positive value because p^2=x

x=149.85  

p^2=149.85

thus p=\pm 12.24

q=\pm 4.90

thus,

\sqrt{119+120 i}=\pm (12.24+i 4.90)

8 0
2 years ago
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