Ask question

Ask question

All categories

3 years ago

5

Zach is 3 years older than twice his sister Michelle's age. Let m represent Michelle's age. Write an expression in terms of m to

represent Zach's age.

2 answers:

dsp733 years ago

4 0

Zach is 3 yrs older then twice his sister (m)

zack is : 2m + 3

Snowcat [4.5K]3 years ago

4 0

2m+3 is the answer to that

You might be interested in

Qiang is inviting 4 friends to a party each friend wants 4 cookies each box has 10 cookies how many boxes should qiang get

Goryan [66]

2
4 friends and each wants 4 so total cookies
required is 16. 1 box has 10 cooking which is not enough,
so you need to get 2 boxes to have more than
16 cookies

3 0

2 years ago

alan drew a polygon with four sides and four angle. All four sides are equal. None of the angle are right angle. What figure did

Nastasia [14]

Alan drew a quadrilateral that is a parallelogram and a rhombus.

4 0

3 years ago

Joshua uses a rule to write the following sequence of numbers 1/6,1/2,5/6,----------,11/2 What rule did Joshua use? What is the

Pani-rosa [81]

Answer: The missing number in the sequence is $\frac{7}{6}$

Step-by-step explanation:

Since we have given that

$\frac{1}{6},\frac{1}{2},\frac{5}{6},----------,\frac{11}{2}$

First term = a= $\frac{1}{6}$

Common difference = d is given by

$d=a_2-a_1\\\\\frac{1}{2}-\frac{1}{6}\\\\=\frac{6-2}{12}\\\\=\frac{4}{12}=\frac{1}{3}\\\\Similarly,\\d=a_3-a_2\\\\=\frac{5}{6}-\frac{1}{2}\\\\=\frac{10-6}{12}\\\\=\frac{4}{12}\\\\=\frac{1}{3}$

Therefore, it forms an arithmetic sequence.

Since, $a_4$ is missing,

So,

$a_4=a+3d\\\\a_4=\frac{1}{6}+3\times \frac{1}{3}\\\\a_4=\frac{1}{6}+1\\\\a_4=\frac{1+6}{6}\\\\a_4=\frac{7}{6}$

Hence, the missing number in the sequence is $\frac{7}{6}$

4 0

3 years ago

Read 2 more answers

Andru [333]

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

◉ $\large\bm{ -4}$

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

Before performing any calculation it's good to recall a few properties of integrals:

$\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}$

$\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }$

So we apply the first property in the first expression given by the question:

$\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}$

And we solve the second integral:

$\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }$

Then we take the last equation and we subtract 10 from both sides:

$\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}$

And we divide both sides by 2:

$\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}$

Then we apply the second property to this integral:

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}$

Then we use the other equality in the question and we get:

$\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}$

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}$

We substract 8 from both sides:

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}$

• $\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}$

7 0

1 year ago

Can someone help me with these 3 plzzzz

aivan3 [116]

1 is 110. 2 is 65 and 3 is 35. You have to double 55 for the first question. For 2 you divide 2 from 130. And the third one shows the angle.

3 0

2 years ago