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ratelena [41]
3 years ago
8

Im confused is 3a the same as 5a squared or 4b to 4b squared​

Mathematics
1 answer:
NeX [460]3 years ago
6 0

Step-by-step explanation:

In order to be like terms, the variables and exponents must be the same.

3a can be combined with 14a and 4a.

4b can be combined with 3b and 16b.

a² cannot be combined with any of the terms.

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Can someone please help me with my homework, thanks!
Damm [24]

Answer: The answer is H!

Step-by-step explanation: H!

7 0
3 years ago
Read 2 more answers
The vertices of ΔMNO are M (1, 3), N (4, 9), and O (7, 3). The vertices of ΔPQR are P (3, 0), Q (4, 2), and R (5, 0). Which conc
vodka [1.7K]

Answer:

They are similar by the definition of similarity in terms of a dilation.

Step-by-step explanation:

Given:

The vertices of ΔMNO are M (1, 3), N (4, 9), and O (7, 3). The vertices of ΔPQR are P (3, 0), Q (4, 2), and R (5, 0).

To choose: the correct option

Solution:

According to distance formula distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For ΔMNO,

MN=\sqrt{(4-1)^2+(9-3)^2}=\sqrt{(3)^2+(6)^2}=\sqrt{45}\\NO=\sqrt{(7-4)^2+(3-9)^2}=\sqrt{(3)^2+(-6)^2}=\sqrt{45}\\\\MO=\sqrt{(7-1)^2+(3-3)^2}=\sqrt{(6)^2+(0)^2}=\sqrt{36}=6\\

For ΔPQR,

PQ=\sqrt{(4-3)^2+(2-0)^2}=\sqrt{(1)^2+(2)^2}=\sqrt{5}\\QR=\sqrt{(5-4)^2+(0-2)^2}=\sqrt{(1)^2+(-2)^2}=\sqrt{5}\\\\PR=\sqrt{(5-3)^2+(0-0)^2}=\sqrt{(2)^2+(0)^2}=2\\

So,

\frac{MN}{PQ}=\frac{\sqrt{45} }{\sqrt{5} }=\sqrt{9}=3\\ \frac{NO}{QR}=\frac{\sqrt{45} }{\sqrt{5} }=\sqrt{9}=3\\\frac{MO}{PR}=\frac{6 }{2 }=3\\

Therefore, They are similar by the definition of similarity in terms of a dilation.

Two triangles are said to be similar if their sides are proportional.

8 0
3 years ago
Given |u| = 10 at ∠135° and |v| = 5 at ∠30°, what expression can be used to find |u + v|?
astraxan [27]

Answer:

B, 10^{2} + 5^{2}- 2(10)(5)cos(75)

Step-by-step explanation:

Using the law of cosines, which is c^{2}=a^{2}+b^{2}-2ab cos(C), you can simply insert all the values.

c=|u+v|

a= r value of u (10)

b= r value of v (5)

To find  C, you simply have to subtract v from u, and then subtract that number from 180 to find the reference angle.

I.E.: ∠135 - ∠30 = ∠105 ↔ 180 - 105 = 75 =  C

so, the completed equation would be 10^{2} + 5^{2}- 2(10)(5)cos(75)

5 0
3 years ago
Read 2 more answers
Write a cosine function that has a midline of 2, an amplitude of 4 and a period of 4/7
ella [17]

Answer:

y= 4cos(\frac{7\pi }{2}x)+2

Step-by-step explanation:

We know that the transformations of a cosine equation can be shown as:

y=±a(b(x-h))+k

Where 'a' is the amplitude

'b' is the horizontal change (Do 2π/b to find the period)

'h' is the horizontal shift

and 'k' is the vertical shift or midline.

------------------------------------------------------

If the amplitude is 4, we can assume a=4.

Since the period is 4/7, we can solve for the 'b' value by:

2\pi /\frac{4}{7}= \frac{7\pi }{2}

Next, since the midline is 2, we know that a vertical shift of 2 occurred. Thus, the 'k' value is 2.

Writing this equation gives us:

y= 4cos(\frac{7\pi }{2}x)+2

4 0
4 years ago
The Northwest High School senior class decided to host a raffle to raise money for their senior trip. They charged $2 for each s
prisoha [69]
Let x be the tickets that the adults bought, andy be the tickets that children bought.
From the problem we have the system of linear equations:
\left \{ {{2x+5y=2686} \atop {y=3x}} \right.

The first thing we are going to do to solve our system, is replacing equation (2) in equation (1), and  then, solve for x
2x+5y=2686
2x+5(3x)=2686
2x+15x=2686
17x=2686
x= \frac{2686}{17}
x=158

Now that we have the number of tickets that the adults bought, lets replace that value in equation (2):
y=3x
y=3(158)
y=474

Last but not least, to find the total number of tickets, we are going to add x and y:
158+474=632

We can conclude that <span>Northwest High School's senior class sold 632 raffle tickets.</span>
7 0
3 years ago
Read 2 more answers
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