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3 years ago

How can I solve this problem?

Mathematics

1 answer:

HACTEHA [7]3 years ago

7 0

Answer:

25*pi or

78.5 square units.

Step-by-step explanation:

If there was no shaded area, the area of the circle would be pi * r^2

r = 10

Area = pi * 10^2

Area = 100 * pi

Since there is a shaded area, the shaded part looks to be 1/4 of the whole. There's no indication of what it actually is, but 1/4 should be close enough.

Area_shade = 1/4 (100*pi)

Area_shade = 25 * pi

That's one answer.

Another is 25 * 3.14 = 78.5 square units

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I haven’t really understood this yet

kakasveta [241]

A, F, and G.

The others have breaks in the range or don’t include all real numbers.

3 0

3 years ago

I'll mark as brainliest if it lets me;-;<br>please answer B. & C. with solution:)

san4es73 [151]

Answer:

B. AC = 22, BC = 22, AB = 44

C. AC = 30, BC = 30, AB = 60

Step-by-step explanation:

$radius \: \overline {OR} \perp chord \: \overline {AB}$ at C. (given)

$\therefore AC = BC$ (perpendicular dropped from the center of the circle to the chord bisects the chord)

$\therefore 2x - 6 = x + 8$

$\therefore 2x - x= 6 + 8$

$\therefore x= 14$

$AC = 2x-6=2(14)-6=28-6= 22$

$BC= x + 8=14 + 8= 22$

AB = AC + BC = 22 + 22 = 44

$radius \: \overline {OR} \perp chord \: \overline {AB}$ at C. (given)

$\therefore BC = AC$ (perpendicular dropped from the center of the circle to the chord bisects the chord)

$\therefore 7x-5=4x+10$

$\therefore 7x - 4x= 5 + 10$

$\therefore 3x= 15$

$\therefore x= \frac{15}{3}$

$\therefore x= 5$

$AC = 4x+10=4(5)+10=20+10= 30$

$BC= 7x - 5=7(5)-5=35-5= 30$

AB = AC + BC = 30 + 30 = 60

7 0

3 years ago

Find the measure of angle 1

Lubov Fominskaja [6]

In ∆FDH, there are two slash marks in two of its legs. This indicates that this triangle is isosceles. If a triangle is isosceles, then it will have two congruent sides and therefore have two congruent angles.

In ∆FDH, angle D is already given to us as the measure of 80°. We can find out the measure of the other angles of this triangle by using the equation:

80 + 2x = 180

Subtract 80 from both sides of the equation.

2x = 100

Divide both sides by 2.

x = 50

This means that angle F and angle H in ∆FDH both measure 50°.

Now, moving over to the next smaller triangle in the picture is ∆DHG. In this triangle, there are also two legs that are congruent which once again indicates that this triangle is isosceles.

First, we have to solve for angle DHG and we do that by using the information obtained from solving for the angles of the other triangle.

**In geometry, remember that two or more consecutive angles that form a line will always be supplementary; the angles add up to 180°.**

In this case angle DHF and angle DHG are consecutive angles which form a linear pair. So, we can use the equation:

Angle DHF + Angle DHG = 180°

50° + Angle DHG = 180°.

Angle DHG = 130°.

Now that we know the measure of one angle in ∆DHG, we can use the same method as the previous step for solving the missing angles. Use the equation:

130 + 2x = 180

2x = 50

x = 25

The other two missing angles of ∆DHG are 25°. This means that the measure of angle 1 is also 25°.

Solution: 25°

8 0

4 years ago

Read 2 more answers

The spinner below is divided into 5 equal sections. What is the probability of landing on a red or green section?

Artyom0805 [142]

Fraction: 2/5

Percentage: 40%

Decimal: 0.4

6 0

3 years ago

Prove that

Pani-rosa [81]

Let's start from what we know.

$(1)\qquad\sum\limits_{k=1}^n1=\underbrace{1+1+\ldots+1}_{n}=n\cdot 1=n\\\\\\
(2)\qquad\sum\limits_{k=1}^nk=1+2+3+\ldots+n=\dfrac{n(n+1)}{2}\quad\text{(arithmetic series)}\\\\\\
(3)\qquad\sum\limits_{k=1}^nk\ \textgreater \ 0\quad\implies\quad\left|\sum\limits_{k=1}^nk\right|=\sum\limits_{k=1}^nk$

Note that:

$\sum\limits_{k=1}^n(-1)^k\cdot k^2=(-1)^1\cdot1^2+(-1)^2\cdot2^2+(-1)^3\cdot3^2+\dots+(-1)^n\cdot n^2=\\\\\\=-1^2+2^2-3^2+4^2-5^2+\dots\pm n^2$

(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first $S_n^+$ with only positive trems (squares of even numbers) and second $S_n^-$ with negative (squares of odd numbers). So:

$\sum\limits_{k=1}^n(-1)^k\cdot k^2=S_n^+-S_n^-$

And now the proof.

1) n is even.

In this case, both $S_n^+$ and $S_n^-$ have $\dfrac{n}{2}$ terms. For example if n=8 then:

$S_8^+=\underbrace{2^2+4^2+6^2+8^2}_{\frac{8}{2}=4}\qquad\text{(even numbers)}\\\\\\
S_8^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{8}{2}=4}\qquad\text{(odd numbers)}\\\\\\$

Generally, there will be:

$S_n^+=\sum\limits_{k=1}^\frac{n}{2}(2k)^2\\\\\\S_n^-=\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\\\\\\$

Now, calculate our sum:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=
\left|\sum\limits_{k=1}^\frac{n}{2}(2k)^2-\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\right|=\\\\\\=
\left|\sum\limits_{k=1}^\frac{n}{2}4k^2-\sum\limits_{k=1}^\frac{n}{2}\left(4k^2-4k+1\right)\right|=\\\\\\$

$=\left|4\sum\limits_{k=1}^\frac{n}{2}k^2-4\sum\limits_{k=1}^\frac{n}{2}k^2+4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|=\left|4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|\stackrel{(1),(2)}{=}\\\\\\=
\left|4\dfrac{\frac{n}{2}(\frac{n}{2}+1)}{2}-\dfrac{n}{2}\right|=\left|2\cdot\dfrac{n}{2}\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\left|n\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\\\\\\
$

$=\left|\dfrac{n^2}{2}+n-\dfrac{n}{2}\right|=\left|\dfrac{n^2}{2}+\dfrac{n}{2}\right|=\left|\dfrac{n^2+n}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\\\\\\\stackrel{(2)}{=}
\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk$

So in this case we prove, that:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk$

2) n is odd.

Here, $S_n^-$ has more terms than $S_n^+$ . For example if n=7 then:

$S_7^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{n+1}{2}=\frac{7+1}{2}=4}\\\\\\
S_7^+=\underbrace{2^2+4^4+6^2}_{\frac{n+1}{2}-1=\frac{7+1}{2}-1=3}\\\\\\$

So there is $\dfrac{n+1}{2}$ terms in $S_n^-$ , $\dfrac{n+1}{2}-1$ terms in $S_n^+$ and:

$S_n^+=\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2\\\\\\
S_n^-=\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2$

Now, we can calculate our sum:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=
\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2-\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2\right|=\\\\\\=
\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}\left(4k^2-4k+1\right)\right|=\\\\\\=
\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}4k^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\$

$=\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-4\left(\dfrac{n+1}{2}\right)^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\=
\left|-4\left(\dfrac{n+1}{2}\right)^2+4\sum\limits_{k=1}^{\frac{n+1}{2}}k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|\stackrel{(1),(2)}{=}\\\\\\
\stackrel{(1),(2)}{=}\left|-4\dfrac{n^2+2n+1}{4}+4\dfrac{\frac{n+1}{2}\left(\frac{n+1}{2}+1\right)}{2}-\dfrac{n+1}{2}\right|=\\\\\\$

$=\left|-n^2-2n-1+2\cdot\dfrac{n+1}{2}\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-2n-1+(n+1)\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-2n-1+\dfrac{(n+1)^2}{2}+n+1-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-n+\dfrac{n^2+2n+1}{2}-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-n+\dfrac{n^2}{2}+n+\dfrac{1}{2}-\dfrac{n}{2}-\dfrac{1}{2}\right|=\left|-\dfrac{n^2}{2}-\dfrac{n}{2}\right|=\left|-\dfrac{n^2+n}{2}\right|=\\\\\\$

$=\left|-\dfrac{n(n+1)}{2}\right|=|-1|\cdot\left|\dfrac{n(n+1)}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk$

We consider all possible n so we prove that:

$\forall_{n\in\mathbb{N}}\quad\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk$

7 0

3 years ago