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Ludmilka [50]
3 years ago
7

Can you guys help me with this question?

Mathematics
1 answer:
IRINA_888 [86]3 years ago
5 0
So below m in the chart, is shows what m equals for you to substitute in the equation

So the second blank would be 42.
Since 7*6=42.

The third blank would equal to 56.
Since 7*8=56
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The store clerk arranges 24 toothbrushes into 4 equal rows. how many toothbrushes are in each row?
Digiron [165]

Answer:

6

Step-by-step explanation:

24/4 = 6

4 0
3 years ago
Read 2 more answers
Find the value of x in the triangle shown below
Firdavs [7]

Answer:

dfbsdfbdfbsdf

Step-by-step explanation:

dfbsdfbfhntjtjnygyhyjnyjaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

7 0
3 years ago
Please help, math question.
Veseljchak [2.6K]
Part a) 
Identify the variable and categories


By reading the above statement, we can find that there are two variables and each variable has 2 categories. The variables and their categories are:
1) Class Number
This variable is further divided into 2 categories i.e. Class 9th and Class 10th.

2) Participation in Extracurricular activities
This variable is further divided into 2 categories based on if students participated or not.

Part b)
Set up and fill 2 way table.

Since, we have 2 variables and each variable has 2 categories the number of data cells will be 2 x 2 = 4 cells

There are total 100 students. 40 students are in class 10th. This means 60 students are in class 9th.Out of 40 students in class 10th, 18 students participate in at least one extracurricular activity. So remaining 22 students do not participate. 32 students from class 9th participate in extracurricular activities, this means the remaining 28 students do not participate. Based on this data, we can fill up the table as shown below
 

6 0
3 years ago
Someone please help me with this lol… have no idea what I’m doing
Sholpan [36]

Given:

\cos \theta =\dfrac{3}{5}

\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

\sin \theta

Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

\sin^2\theta=1-\cos^2\theta

\sin \theta=\pm \sqrt{1-\cos^2\theta}

It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

After substituting \cos \theta =\dfrac{3}{5}, we get

\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

\sin \theta=-\sqrt{1-\dfrac{9}{25}}

\sin \theta=-\sqrt{\dfrac{25-9}{25}}

\sin \theta=-\sqrt{\dfrac{16}{25}}

\sin \theta=-\dfrac{4}{5}

Therefore, the correct option is B.

6 0
2 years ago
Solve the quadratic equation.<br> 2x2 + 35 = 17x
atroni [7]
Get it to equal 0

minus 17x from both sides

2x²-17x+35=0

tricky part
get something like
(2x-a)(x-b)
we know the signs  are negative since the middle term is negative and last term is positive

so a and b are either 1 and 35 or 5 and 7

(2x-1)(x-35), nope
(2x-35)(x-1) nope
(2x-5)(x-7), nope
(2x-7)(x-5), yep

(2x-7)(x-5)=0
set to zero
2x-7=0
2x=7
x=7/2

x-5=0
x=5


x=7/2 and 5
8 0
2 years ago
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