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2 years ago

Need help if anybody can help.

Mathematics

1 answer:

Darina [25.2K]2 years ago

8 0

Answer:

46.5

Step-by-step explanation:

Of all straight angles have 180 degrees in them.

So, to find x we need to form this equation:

x + 65 + x - 22 = 180

2x + 43 = 180

2x = 180-43

2x = 137

68.5 degrees = x

68.5-22 = 46.5 degrees.

m<CBD = 46.5 degrees.

m<GBF = also 46.5 degrees because it is a vertical angle to <46.5 degrees.

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5m - 3n = 19; m - 6n = -7 /pls. help me solve this linear equation with two variables.

Allisa [31]

Answer:

m = 5, n = 2

Step-by-step explanation:

$5m - 3 n = 19\\m - 6n = - 7 \ => m = -7 + 6n\\\\Substitute \ m \ in \ 5m - 3n = 19\\\\5(-7 + 6n) - 3n = 19\\-35 + 30n -3n = 19\\27n = 19 + 35 \\27n = 54 \\n = 2\\\\Substitute \ n = 2 \ in \ m = -7 + 6n\\\\m = -7 + (6 \times 2) = -7 + 12 = 5$

5 0

3 years ago

I need help with 2,3 and 6! Please Help!

Georgia [21]

2. If you already know Faulhaber's formula, which says

$\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6$

then it's just a matter of setting $n=4$ . If you don't, then you can prove that it works (via induction), or compute the sum by some other means. Presumably you're not expected to use brute force and just add the squares of 1 through 4.

Just to demonstrate one possible method of verifying the formula, suppose we start from the binomial expansion of $(k-1)^3$ , do some manipulation, then sum over $1\le k\le n$ :

$(k-1)^3=k^3-3k^2+3k-1$
$\implies k^3-(k-1)^3=3k^2-3k+1$
$\implies\displaystyle\sum_{k=1}^n(k^3-(k-1)^3)=\sum_{k=1}^n(3k^2-3k+1)$

The left side is a telescoping series - several terms in consecutive terms of the series will cancel - and reduces to $n^3$ . For example,

$\displaystyle\sum_{k=3}^2(k^3-(k-1)^3)=(1^3-0^3)+(2^3-1^3)+(3^3-2^3)=3^3$

Distributing the sum on the right side across each term and pull out constant factors to get

$\displaystyle n^3=3\sum_{k=1}^nk^2-3\sum_{k=1}^nk+\sum_{k=1}^n1$

If you don't know the formula for $\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$ , you can use a similar trick with the binomial expansion $(k-1)^2$ , or a simpler trick due to Gauss, or other methods. I'll assume you know it to save space for the other parts of your question. We then have

$\displaystyle n^3=3\sum_{k=1}^nk^2-\frac{3n(n+1)}2+n$
$\implies\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6$

and when $n=4$ we get 30.

3. Each term in the sum is a cube, but the sign changes. Recall that $(-1)^n$ is either 1 if $n$ is even or -1 if $n$ is odd. So we can write

$1^3-2^3+3^3-4^3+5^3=\displaystyle\sum_{k=1}^5(-1)^{k-1}k^3$

( $k+1$ as the exponent to -1 also works)

6. If $0\le k\le n$ , and $i=k+1$ , then we would get $1\le k+1\le n+1\iff1\le i\le n+1$ . So the sum with respect to $i$ is

$\displaystyle\sum_{k=0}^n\frac{k^2}{k+n}=\sum_{i=1}^{n+1}\frac{(i-1)^2}{i+n-1}$

4 0

3 years ago

For a class trip, the teachers would like to have one adult for every ten students. There are 190 students on the trip. How many

coldgirl [10]

You can solve this problem with the following

$\frac{1}{10} = \frac{a}{190}$
$10a = 190$
$a = 19$

There should be 19 adults on the trip.

6 0

3 years ago

NEED ASAP!!!! Assessment items Select from the drop-down menu to correctly identify the property shown. (−3.4+(−1.5))+6.2=−(3.4

Vikki [24]

Commutative property.

4 0

3 years ago

The image of (6, 9) under a dilation is (4, 6).

Ad libitum [116K]

For this case we have the following ordered pairs:
(6, 9)
(4, 6)
The scale factor can be found in two different ways:
$k = 6/4 = 3/2 = 1.5

k = 9/6 = 3/2 = 1.5$
Therefore, the scale factor is given by:
$k = 1.5
$
Answer:
The scale factor of the dilation is 1.5

8 0

3 years ago

Read 2 more answers

Need help if anybody can help.​

Need help if anybody can help.