Describe strategies for finding factors or multiples of a number

Mathematics

1 answer:

Darina [25.2K]3 years ago

3 0

Say if you wanted to find factors of 6, the you'd multiply that number by any number to find a factor or multiple.

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Solve for d:<br> d+(−5.004)=2.826 <br> Include explanation on how you got your answer :)

aleksklad [387]

Answer:

$d = 7 .83$

Step-by-step explanation:

1. Remove parentheses.

$d + - 5.004 = 2.826$

2. Simplify d + -5.004 to d - 5.004.

$d - 5.004 = 5.004$

3. Add 5.004 to both sides.

$d = 2.826 + 5.004$

4. simplify 2.826 + 5.004 to 7.83.

$d = 7.83$

Therefor, rhe answer is D = 7.83

8 0

3 years ago

The least common multiple of two numbers is 40. The difference of the two numbers is 2.What is the value of the greater of the t

Natalka [10]

20=2*2*5 42=2*3*7 that is your awnser

6 0

3 years ago

Prove: cot(x) sec^4(x) = cot(x) +2tan(x) + tan^3(x)

masha68 [24]

$\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad\qquad 
sec(\theta)=\cfrac{1}{cos(\theta)}\quad\qquad 1+tan^2(\theta)=sec^2(\theta)\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\\\\
-------------------------------\\\\
cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)\\\\
-------------------------------\\\\$

$\bf \textit{so, let's do the left-hand-side}\\\\
cot(x)sec^2(x)sec^2(x)\implies cot(x)[1+tan^2(x)][1+tan^2(x)]
\\\\\\
cot(x)[1^2+2tan^2(x)+tan^4(x)]
\\\\\\
cot(x)+2tan^2(x)cot(x)+tan^4(x)cot(x)
\\\\\\
cot(x)+2\cdot \cfrac{sin^2(x)}{cos^2(x)}\cdot \cfrac{cos(x)}{sin(x)}+\cfrac{sin^4(x)}{cos^4(x)}\cdot \cfrac{cos(x)}{sin(x)}
\\\\\\
cot(x)+2\cdot \cfrac{sin(x)}{cos(x)}+\cfrac{sin^3(x)}{cos^3(x)}\implies \boxed{cot(x)+2tan(x)+tan^3(x)}$

8 0

3 years ago

Find the quadratic polynomial if zeros of it are 2 and 1/3 respectively

muminat

Answer:

$\boxed{\sf \ \ \ 3x^2-7x+2 \ \ \ }$