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tia_tia [17]
3 years ago
13

What is the area of the following circle R=7

Mathematics
2 answers:
Roman55 [17]3 years ago
6 0

Answer:

Area of the circle = 153.86\,units^2

Step-by-step explanation:

Area of the Circle= \pi \times r^2

For, r=7 \,units

As,

\pi =\dfrac{22}{7} =3.14

Area is:

            =3.14\times 7 \times 7\\\\=3.14\times 49\\\\=153.86\,units^2

The area of the circle is : 153.86\,units^2

In-s [12.5K]3 years ago
5 0

Answer:

153.86

Step-by-step explanation:

7^2 x 3.14

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What is the solution to the equation -2(1+5a)=-6(a+3) ?
Lilit [14]


The solution is a=4

-2(1+5a)=-6(a+3)

-2-10a=-6a-18

+18              +18

16-10a=-6a

     +10a      +10a

16=4a

divide by 4

4=a


7 0
3 years ago
If x is increased by 20%.<br>then what<br>will be the increased percent if there will be x^2?<br>​
nevsk [136]

Answer:

the area will be increased by 44%

Step-by-step explanation:

5 0
3 years ago
Work out 1 3/4 x 4 1/6
sattari [20]

Convert mixed number to improper fraction:

1\dfrac{3}{4}=\dfrac{1\cdot4+3}{4}=\dfrac{4+3}{4}=\dfrac{7}{4}\\\\4\dfrac{1}{6}=\dfrac{4\cdot6+1}{6}=\dfrac{24+1}{6}=\dfrac{25}{6}\\\\1\dfrac{3}{4}\times4\dfrac{1}{6}=\dfrac{7}{4}\times\dfrac{25}{6}=\dfrac{7\times25}{4\times6}=\dfrac{175}{24}=\dfrac{168+7}{24}=7\dfrac{7}{24}

8 0
3 years ago
I need help , I don’t understand this
marta [7]
#2. First, we factor each polynomial. Then, if any terms on both the top and the bottom of the fraction match, they cancel out. So... we do just that. You end up with:

\frac{x(x-4)}{(x+9)(x-4)}

Notice there's an (x-4) on both top and bottom. So they cancel out. That leaves us with your answer of \frac{x}{(x+9)}

#3. We do the same thing as above then multiply and simplify. In the interest of space, I'll cut straight to some simplification. 

\frac{2(x+2)^{3} }{6x(x+2)} ( \frac{5}{(x-2)^{2} } )

Now we start cancelling. For the first fraction, there are 3 (x+2)'s on top and 1 on the bottom so we will cancel out the one on the bottom and leave 2 (x+2)'s on top. There are no more polynomials to cancel out so now we multiply across:

\frac{10(x+2)^{2} }{6x(x-2)^{2} }

10 and 6 share a GCF of 2 so we divide both of those by 2. This leaves us with the final answer of:

\frac{5(x+2)^{2} }{3x(x-2)^{2} }

#4. This equation introduces division and because of it, we must flip the second fraction to make the division sign into a multiplication symbol. Again for space, I'll flip the fraction and simplify in one step. 

\frac{3(x+2)(x-2)}{(x+4)(x-2)} ( \frac{x+4}{6(x+3)})

Now we do our cancelling. First fraction has (x - 2) in the top and bottom. They're gone. The first fraction has a (x + 4) on the bottom and the second fraction has one on the top. Those will also cancel. This leaves you with:

\frac{3(x+2)}{6(x+3)}

3 and 6 share a GCF of 3 so we divide both numbers by this. This leaves you with your final answer:

\frac{x+2}{2(x+3)}

#5. We are adding so we first factor both fractions and see what we need to multiply by to make the denominators the same. I'll do the former first. (10 - x) and (x - 10) are not the same so we multiply the first equation (top and bottom) by (x - 10) and the second equation by (10 - x). Because they will now have the same denominator we can combine them already. This gives us:

\frac{(3+2x)(x-10)+(13+x)(10-x)}{(10-x)(x-10)}

Now we FOIL each to expand and then simplify by combining like terms. Again for space, I'm just showing the result of this; you end up with:

\frac{x^{2}-20x+100}{(10-x)(x-10)}

Now we factor the top. This gives you 2 (x - 10)'s on top and one on bottom. So we just leave one on the top and cancel the bottom one out. This leaves you with your answer:

\frac{x+10}{10-x}

#6. Same process for this one so I won't repeat. I'll just show the work.

\frac{3}{(x-3)(x+2)} +  \frac{2}{(x-3)(x-2)} becomes

\frac{3(x-2) + 2(x+2)}{(x-3)(x+2)(x-2)} which equals

\frac{3x - 6 + 2x + 4}{(x-3)(x+2)(x-2)} giving you the final answer

\frac{5x - 2}{(x-3)(x+2)(x-2)}

#7. For this question we find the least common denominator to make the denominators match. For 5, x, and 2x, the LCD is 10x. So we multiply top and bottom of each fraction by what would make the bottom equal 10x. This rewrites the fraction as:

\frac{3x}{5} ( \frac{2x}{2x}) * ( \frac{5}{x}( \frac{10}{10}) -  \frac{5}{2x} ( \frac{5}{5}))

Simplify to get:

\frac{3x}{5}  * ( \frac{25}{10x})

After simplifying again, you end up with your final answer: 

\frac{3}{2}




8 0
3 years ago
F(x)=2x^2-3x+7 evaluate f(-2)
Vilka [71]

Answer:

Step-by-step explanation:

Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?

Let  p(x)=kpx+dp  and  q(x)=kqx+dq  than

f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7  

p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7  

(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)  

So you want:

−2kpk2q=0  

and

kpkq=−1  

and

2kpd2p−3kpdq+7=0  

Now I amfraid this doesn’t work as  −2kpk2q=0  that either  kp  or  kq  is zero but than their product can’t be anything but  0  not  −1 .

Answer: there are no such linear functions.

6 0
2 years ago
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