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3 years ago

Two more than 16 times a number is the same as 21 less than the product of -7 and the number

Mathematics

1 answer:

erastova [34]3 years ago

6 0

16x+2=-7x-21

x = -1

So yeah

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The ratio of two numbers is 1:4, and the sum of these numbers is 40. Find the numbers

Irina-Kira [14]

X + 4x = 40

5x = 40

5x/5 = 40/5

x = 8

1x 8 = 8

4x 8 = 32

Check=

8 + 32 = 40

The numbers are 8 and 32

3 0

2 years ago

This graph represents a transformation of the parent cube root function. Replace the values of h and k to create the equation of

Zigmanuir [339]

Answer: H is 5 and K is 2

Step-by-step explanation:

8 0

3 years ago

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velikii [3]

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2 years ago

Use the multiplication principle to tell the total number of three-card hands (permutations) that can be made with two aces

alina1380 [7]

We are asked for the total or three card hands that can be made with two aces. The three-card hand should have two aces. We will use permutation in solving this problem and the solution is shown below:
nPr = n! /(n-r)!
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r=2
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The answer is 6.

3 0

3 years ago

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

8 0

2 years ago