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2 years ago

How do you solve this equation thankyou. 5n+34=-2(1-7n)

Mathematics

1 answer:

JulijaS [17]2 years ago

5 0

$5n+34=-2(1-7n) \\
5n+34=-2+14n\\
-9n=-36\\
n=4$

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HELP ME PLEASE solve the right triangle

Amiraneli [1.4K]

ST = 7.2

RT = 4.04

m∠S = 34°

Solution:

Given ΔRST is a right triangle.

RS = 6, ∠T = 56°

$$\csc A=\frac{\text {Hypotenuse} }{\text {Opposite side}}$

$$\csc A=\frac{ST}{RS}$

$$\csc 56^\circ=\frac{ST}{6}$

$$\csc 56^\circ \times 6=ST$

1.206 × 6 = ST

7.236 = ST

ST = 7.2

$$\cot A=\frac{\text {Adjacent side} }{\text {Opposite side}}$

$$\cot A=\frac{RT}{RS}$

$$\cot 56^\circ=\frac{RT}{6}$

$$\cot 56^\circ \times 6=RT$

0.674 × 6 = RT

4.044 = RT

RT = 4.04

To find the measure of angle S.

Sum of the interior angles of the triangle = 180°

m∠R + m∠S + m∠T = 180°

90° + m∠S + 56° = 180°

m∠S + 146° = 180°

m∠S = 180° – 146°

m∠S = 34°

8 0

2 years ago

10. Match each measurement with the correct number of dimensions.

4vir4ik [10]

1. area - a
2. height - b
3. length - b
4. volume - c

8 0

2 years ago

Unit 7 homework 5 Gina Wilson

OlgaM077 [116]

Huh??? Where is the question?

4 0

2 years ago

Read 2 more answers

A pack of 6 cans of pop costs £1.80. A pack of 10 cans of pop costs £3.50.

Lelechka [254]

Let us look at the cost per can of both packs.

Cost of can in 6-pack = 1.80/6

= £0.30

Cost of can in 10-pack = 3.50/10

= £0.35

Since the cost of a can in the 6-pack is cheaper, it would be better to buy the 6-pack if you were looking to save money.

5 0

2 years ago

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

2 years ago