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4 years ago

13

What is the prime factorization of 154? 2 × 7 × 11 2 × 7 × 13 2 × 5 × 11 2 × 5 × 13

1 answer:

ruslelena [56]4 years ago

4 0

The answer is....<span>2 × 7 × 11.</span>

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I will mark brainliest if correct - Maria solved an equation as shown below.

zepelin [54]

Hello!

Since x = 19 the answer to the problem is 19

Hope this helps!

8 0

4 years ago

Read 2 more answers

Find the area of the shape shown below.

hichkok12 [17]

Answer:

26.75 units ^2

Step-by-step explanation:

Since the shape is complex, divide it into 3 right angled triangles and one square. Find the area of these individual shapes first, then fin the sum of these area to calculate the ultimate area of e complex shape:

Triangle 1 = 1/2 x 2 x 5 = 5 units ^2

Triangle 2 = 1/2 x 2 x 2 = 2 units ^2

Triangle 3 = 1/2 x 3.5 x 9 = 15.75 units ^2

Square = 2 x 2 = 4 units squared.

Now add all these up 15.75 + 2 + 5 + 4 = 26.75 units squared.

Hope this helps

6 0

3 years ago

6-4x=6x-8x+2<img src="https://tex.z-dn.net/?f=6-4x%3D6x-8x%2B2" id="TexFormula1" title="6-4x=6x-8x+2" alt="6-4x=6x-8x+2" align="

e-lub [12.9K]

$6-4x=6x-8x+2 \\
6-4x=-2x+2 \\
-4x+2x=2-6 \\
-2x=-4 \\
\underline{\underline{x=2}}$

6 0

4 years ago

Use the equation below to answer the question.

JulijaS [17]

Answer:

The signa notation to represent the first five ten f(x) is given by $f(x)=\sum\limits^{5}_{n=1}a_n$

Step-by-step explanation:

Given sequence is -5, -9, 13...

Let f(x) be the given sequence and is denoted by

$f(x)=\{-5,-9,-13,...\}$

Let the first term be $a_1, 2^{\textrm{nd}}$ term be $a_3,...$

ie, $a_1=-5,a_2=-9, a_3=-13,...$

To find the common difference d:

$d=a_2-a_1$

$=-9-(-5)$

$=-9-(+5)$

$d=-4$

$d=a_3-a_2$

$=-13-(-9)$

$=-13+9$

$d=-4$

Therefore the common difference d is -4 for given sequence f(x) with $a_1=-5$ and d=-4, the seqence f(x) is an arithmetic sequence

By defintion of arithmetic sequence

$a_n=a+(n-1)d\hfill(1)$

Now to find $a_4, a_5$ :

put n=4 in equation (1)

$a_4=a+(4-1)d$

$a_4=-5+(3)(-4)$ [since a=-5, d=-4]

$=-5-12$

$a_4=-17$

in equation (1)

$a_5=a+(5-1)d$

$a_5=-5+(4)(-4)$ [since a=-5, d=-4]

$=-5-16$

$a_5=-21$

Therefore $a_4=-17$ and $a_5=-21$

Therefore $f(x)=\{-5,-9,-13,-17,-21,...\}$

Now to represent the sum of the first five terms of f(x) using sigma notation as below

$f(x)=\sum_{n=1}^5 a_n$

where $\sum_{n=1}^5 a_n=a_1+a_2+a_3+a_4+a_5$

$-5-9-13-17-21$

$\sum_{n=1}^5 a_n=-65$

4 0

3 years ago

Is this proportional

stellarik [79]

Answer:

yes

Step-by-step explanation:

its a straight line

3 0

4 years ago