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2 years ago

Help please i’ve never done it with negative numbers so it’s a little more confusing just need explanation

Mathematics

1 answer:

jok3333 [9.3K]2 years ago

8 0

To answer this question, we need to pose the problem algebraically as follows:

1. The sum of two numbers is -29:

$x+y=-29$

2. The difference of the two numbers is -13:

$x-y=-13$

Then, we have a linear system of equations, and we can solve it by elimination, substitution, as the most suitable for this case. Then, we have that we can sum both equations and we get:

Thus, we have that the value for x = -21. We can use the first equation to substitute this value of x to find the value for y. Then, we have:

$x+y=-29\Rightarrow-21+y=-29\Rightarrow y=-29+21\Rightarrow y=-8$

Therefore, the two numbers are -21 and -8.

We can check this as follows:

1. -21 + (-8)= -21 -8 = -29

2.-21 -(-8) = -21 + 8 = -13

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Step-by-step explanation:

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3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

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4 years ago

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kumpel [21]

Answer:

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Step-by-step explanation:

4 seconds : 2 min

however we know that 1 min has 60 seconds. Hence 2 min will have 2 x 60 = 120 seconds

Therefore we can substitute 2 min with 120 seconds

4 seconds : 2 min

4 seconds : 120 seconds (simplify by dividing both sides by 4)

4/4 : 120/4

1 : 30

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3 years ago

Read 2 more answers

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vova2212 [387]

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grin007 [14]

Answer:

B. 0

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3 years ago