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r-ruslan [8.4K]
3 years ago
7

What is x if -10=10/3x

Mathematics
2 answers:
Taya2010 [7]3 years ago
6 0
3*-10=-30 and then -30 divided by 10 gives you negative 10
There fore the answer is
x=-10
Elanso [62]3 years ago
4 0

Answer:

-1/3

Step-by-step explanation:

-10(3x)=10

-30x=10

x=10/-30

x=-1/3

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Find the area of the square below
Mkey [24]

Answer:

25 un²

Step-by-step explanation:

5 · 5 = 25 un²

8 0
3 years ago
What is the range of a function y=|x+3| -3 if the domain is x€R
Anastasy [175]

Answer:

y ≥−3

Step-by-step explanation:

1 By inspecting the graph, the range is:

y\ge -3y≥−3

Done

3 0
3 years ago
I need some help ill give brainly
anastassius [24]

Answer:

Step-by-step explanation:

5 0
4 years ago
PLS HELP SHOW ALL YOUR WORKING OUT BRAINLIEST
forsale [732]

Area of a triangle = height x base ÷ 2

When ever you have an isosceles triangle, remember that you can split it in half to form two right angled triangles - which allow you to use Pythagoras' Theorem.

So we split 10cm in half to get 5cm.

We can then use 13cm and 5cm to work out the height:

Height = √(13²  - 5²)      <em>(Note: you subtract because you are using the    </em>

           = √144              <em>(hypotenuse)</em>

<em>            =  </em>12

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

Now to get the area, we just multiply the base (which is 10) by the height (which is 12) and divide by 2:

Area = 12 x 10 ÷ 2

        = 6 x 10               <em>(note: 12 ÷ 2 = 6 )</em>

        = 60 cm²

_____________________________________

Answer:

60 cm²

8 0
4 years ago
Write the expression for W in terms of K AND write the expression of Z in terms of K PLZ HELP
tekilochka [14]

Answer:

w=100-\frac{(100-k)^2}{100}\\\\z={\frac{100}{\left(1-\frac{k}{100}\right)^2} -100

Step-by-step explanation:

\underline{w \ in \ terms \ of \ z:}

d=\left(1-\frac{k}{100}\right)c\\a=\left(1-\frac{k}{100}\right)d\\ \Rightarrow a=\left(1-\frac{k}{100}\right)^2c...........(1)\\

Again \ a=\left(1-\frac{w}{100}\right)c...........(2)\\

from equation (1) and (2)

\left(1-\frac{k}{100}\right)^2=\left(1-\frac{w}{100}\right)\\\frac{w}{100}=1-\left(1-\frac{k}{100}\right)^2\\w=100-\frac{(100-k)^2}{100}

\underline{z \ in \ terms \of \ w:}

c=\left(1+\frac{z}{100}\right)a.......(3)\\

from equation (1) and (3)

a=\left(1-\frac{k}{100}\right)^2\left(1+\frac{z}{100}\right)a\\\left(1+\frac{z}{100}\right)=\frac{1}{\left(1-\frac{k}{100}\right)^2}\\\frac{z}{100}=\frac{1}{\left(1-\frac{k}{100}\right)^2}-1\\z=100\left[\frac{1}{\left(1-\frac{k}{100}\right)^2}-1\right]

5 0
3 years ago
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