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Harrizon [31]
2 years ago
5

someone is eating a 50 gram chocolate bar which contains 30% cocoa how many grams of cocoa is in the chocolate bar

Mathematics
2 answers:
mixer [17]2 years ago
4 0
It's would be 15 grams. Your ratio is x/50=30/100. You then would cross multiply 30×50 and end up with 1500. After that you would divide by 100 and have your answer, 15g.
DIA [1.3K]2 years ago
3 0
Well , 10% is 5 g so if you times those 5 g by 3 it is 15 g there are 15 grams of cocoa in the chocolate bar
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-3 because you could eliminate the 3x by subtracting it by 3x
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The sum of 4 consecutive even integers is -12. find the largest integer.
Ipatiy [6.2K]

Answer:

The largest even integer is 0

Step-by-step explanation:

Given as :

The sum of four consecutive even number = - 12

Let The first even number = 2 x

The second even number = (2 x + 2)

The third even number = (2 x + 4)

The fourth even number = (2 x + 6)

<u>According to question</u>

sum of four consecutive even number = - 12

Or, 2 x + (2 x + 2) + (2 x + 4) + (2 x + 6) = - 12

Or, (2 x + 2 x + 2 x + 2 x) + (2 + 4 + 6) = - 12

Or, 8 x + 12 = - 12

Or, 8 x = - 12 - 12

Or, 8 x = - 24

∴  x = \frac{- 24}{8}

i.e  x = - 3

now, putting the value of x

The first even number = 2 × - 3 = - 6

The second even number = (2 × - 3 + 2) = - 6 + 2 = - 4

The third even number = (2 × - 3 + 4) = - 6 + 4 = - 2

The fourth even number = (2 × - 3 + 6) = - 6 + 6 = 0

So, The largest even integer = 0

Hence, The largest even integer is 0  Answer

6 0
3 years ago
Classify the polynomial 2x3 + 6x2 - 4 by the number of terms.
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Step-by-step explanation:

8 0
2 years ago
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which of the following correctly describes the end behavior of the polynomial function, f(x)=2x^4-3x^2+2x
inna [77]

Answer:

Both ends go up.


Step-by-step explanation:


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3 years ago
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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
GarryVolchara [31]

Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

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}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

7 0
3 years ago
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