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3 years ago

Solve and check the following equation. 8d + 21 = 12d + 5 d =

Mathematics

1 answer:

olganol [36]3 years ago

3 0

Subtract 8d from both sides

21 = 12d + 5 - 8d

Simplify 12d + 5 - 8d to 4d + 5

21 = 4d + 5

Subtract 5 from both sides

21 - 5 = 4d

Simplify 21 - 5 to 16

16 = 4d

Divide both side by 4

16/4 = d

Simplify 16/4 to 4

4 = d

Switch sides

d = 4

Check answer

8d + 21 =12d + 5

Let d = 4

8 × 4 + 21 = 12 × 4 + 5

Simplify 8 × 4 to 32

32 + 21 = 12 × 4 + 5

Simplify 12 × 4 to 48

32 + 21 = 48 + 5

Simplify 32 + 21 to 53

53 = 48 + 5

Simplify 48 + 5 to 53

53 = 53

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7. Find the sum or difference. -2(d +1) + (6d +4)

Arturiano [62]

Answer:

Step-by-step explanation:

Sum

-2d + 1 + 6d + 4

Solving like terms

4d + 5

Difference

-2d + 1 - 6d - 4

-8d - 3

4 0

3 years ago

Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?

Bogdan [553]

Step-by-step explanation:

$\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}$

$\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}$

Taking sin²θ common in both numerator & denominator, We get :

$\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

$\mathsf{\implies \dfrac{1 -cot^2\theta}{1 + cot^2\theta}}$

$\mathsf{Given :\;cot\theta = \dfrac{1}{\sqrt{5}}}$

$\mathsf{\implies \dfrac{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2}{1 + \left(\dfrac{1}{\sqrt{5}}\right)^2}}$

$\mathsf{\implies \dfrac{1 - \dfrac{1}{5}}{1 + \dfrac{1}{5}}}$

$\mathsf{\implies \dfrac{\dfrac{5 - 1}{5}}{\dfrac{5 + 1}{5}}}$

$\mathsf{\implies \dfrac{5 - 1}{5 + 1}}$

$\mathsf{\implies \dfrac{4}{6}}$

$\mathsf{\implies \dfrac{2}{3}}$

Hence, option (a) 2/3 is your correct answer.

3 0

2 years ago

P, Q, and R are three different points. PQ = 3x + 2, QR = x, RP = x + 2, and . List the angles of PQR in order from largest to

horsena [70]

PQR is a triangle
QR = x ⇒ x>0
If x>0 then:
QR is the smallest side
PQ is the largest side

In the triangle, the largest angle lies opposite the largest side.
The angles of ΔPQR in order from largest to smallest:
∠R is largest [opposite to PQ]
∠Q is middle [opposite to RP]
∠P is smallest [opposite to QR]

4 0

3 years ago

I need help with this one too please !i mark brainliest to everyone

denis-greek [22]

Answer:

I believe A is the answer

5 0

2 years ago

Read 2 more answers

How can you use the measures of center and the range to compare two populations?

DochEvi [55]

The measure of center and range to be used to compare the population is the median and the inter quartile range.

Explanation:

You utilize the median and the interquartile range (IQR) to portray skewed distributions of information. To look at two populaces, utilize the mean and the MAD when the two populations are symmetric. Utilize the middle and the IQR when possibly one or the two circulations are slanted.

In statistics and probability theory, the median is the worth isolating the higher half from the lower half of an information test, a populace or a likelihood circulation. For an informational collection, it might be thought of as the "center" esteem.

8 0

3 years ago