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3 years ago

Will give 25 points and brainliest answer if correct

Mathematics

2 answers:

zimovet [89]3 years ago

8 0

The answer is A, which is 4,3,10,17,24,31.

AleksAgata [21]3 years ago

5 0

The first term a1 is given to be -4. The formula for the next terms: an = an-1 + 7 means to add 7 to the previous term to get the next.

-4, 3, 10, 17, 24, 31

Answer A

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3k^2-11k-4=0 solve with quadratic formula

lawyer [7]

It should be k= 4 or -1/3

8 0

2 years ago

Given the line below. Write the point-slope form of the given line that passes through the points (-4, -1) and (3, -3). Identify

allsm [11]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~{{ -4}} &,&{{ -1}}~) 
% (c,d)
&&(~{{ 3}} &,&{{ -3}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies 
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{-3-(-1)}{3-(-4)}\implies \cfrac{-3+1}{3+4}\implies -\cfrac{2}{7}$

$\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-1)=-\cfrac{2}{7}[x-(-4)]
\\\\\\
y+1=-\cfrac{2}{7}(x+4)$

4 0

3 years ago

What are the first and last terms of this geometric series

11Alexandr11 [23.1K]

ANSWER

E. 10 and 640

EXPLANATION

The given series is

$\sum_{i=1}^45 \times 2^{2i-1}$

To find the first term of the series we substitute i=1 to get,

$a_1 = 5 \times {2}^{2 \times 1 - 1}$

$a_1 = 5 \times {2}^{ 1} = 10$

For the last term of the series, we substitute i=4 to obtain,

$a_4 = 5 \times {2}^{2 \times 4 - 1}$

$a_4 = 5 \times {2}^{7}$

This implies that,

$a_4 = 5 \times 128 = 640$

Therefore the correct answer is E

7 0

3 years ago

∠A and \angle B∠B are supplementary angles. If m\angle A=(2x-16)^{\circ}∠A=(2x−16) ∘ and m\angle B=(5x+28)^{\circ}∠B=(5x+28) ∘ ,

Otrada [13]

Answer:

The measure of angle A is 32

Step-by-step explanation:

A+B=180

2x-16+5x+28=180

7x+12=180

7x=168

x=24

A=2(24)-16

A=48-16

A=32

6 0

3 years ago

TanA+tanB+tanAtanB=1 If A+B=45 degree

djyliett [7]

Answer:

See below

Step-by-step explanation:

$\tan A+\tan B + \tan A.\tan B = 1\\\\ \implies \: \tan A + \tan B = 1 - \tan A.\tan B \\ \\ \implies \: \frac{\tan A + \tan B}{ 1 - \tan A.\tan B } = 1\\ \\ \implies \: \tan (A + B) = 1 \\ \\ \implies \: \tan (A + B) =\tan 45 \degree \\ \\ \implies \: A + B = 45 \degree \\ \\ thus \: proved$

8 0

2 years ago