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murzikaleks [220]
3 years ago
13

Can someone help me

Mathematics
1 answer:
notsponge [240]3 years ago
4 0
36 is the correct answer u believe
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Solve a) 2(x+3)=x-4 b) 4(5x-2)=2(9x+3)
Nat2105 [25]

Answer:

a) -10 b) 7

Step-by-step explanation:

a) 2(x+ 3) = x - 4

2x + 6 = x - 4

2x - x = -6 - 4

x = -10

b) 4(5x - 2) = 2(9x + 3)

20x - 8 = 18x + 6

20x - 18x = 6 + 8

2x = 14

x = 14/2

x = 7

5 0
3 years ago
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What are prime Numbers
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2,3,5,7,11,13,821, and 823


please can i have a brainliest
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3 years ago
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Which of these is equal to 0.13?<br> 13<br> A.<br> 99<br> 12<br> B.<br> 90<br> c. 1
Norma-Jean [14]

Answer:

Is this the inter question

3 0
3 years ago
I need help understanding how to do the problem
notsponge [240]
So.. if you notice the picture below

is really just 3/4 of a cylinder, or, a full cylinder, and then you slice 1/4 off of it
notice the right-angle in your picture at the bottom, is cut at a right-angle, meaning, the cutout is 1/4 of the volume

so \bf \textit{volume of a cylinder}\\\\&#10;V=\pi r^2 h\qquad &#10;\begin{cases}&#10;r=radius\\&#10;h=height&#10;\end{cases}\qquad \frac{3}{4}\textit{ of that}\implies \cfrac{3}{4}\cdot \pi r^2 h

4 0
3 years ago
According to the Fundamental Theorem of Algebra, the graph of f(x) = x2 - 4x + 3, has roots. From the graph we can see that it h
andreyandreev [35.5K]

Answer:

The graph f(x)=x^2-4x+3  has two zeros namely 3 and 1.

Step-by-step explanation:

Consider the given equation of graph f(x)=x^2-4x+3

According to the Fundamental Theorem of Algebra

For a given polynomial of degree n can have a maximum of n roots.

Thus, for the given equation f(x)=x^2-4x+3  the degree of polynomial is 2 , thus the function can have maximum of 2 roots.

We know at roots the value of function is 0 that is f(x) = 0,

Substitute f(x) = 0 , we get, f(x)=x^2-4x+3=0

This is a quadratic equation, x^2-4x+3=0

We first solve it manually and then check by plotting graph.

Quadratic equation can be solved using middle term splitting method,

here, -4x can be written as -x-3x,

x^2-4x+3=0 \Rightarrow x^2-x-3x+3=0

\Rightarrow x(x-1)-3(x-1)=0

\Rightarrow (x-3)(x-1)=0

Using zero product property, a\cdot b=0 \Rightarrow a=0\ or \ b=0

\Rightarrow (x-3)=0 or \Rightarrow (x-1)=0

\Rightarrow x=3 or \Rightarrow x=1

Thus, the two zero of f(x) are 3 and 1.

We can also see on graph attached below that the graph f(x)=x^2-4x+3  has two zeros  namely 3 and 1.  

3 0
3 years ago
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