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2 years ago

Increase 50 kg by 10%

Mathematics

2 answers:

PolarNik [594]2 years ago

5 0

10% of 50kg is 5kg
50kg add 5kg is 55kg
=55kg

Alika [10]2 years ago

5 0

If it decreased ⬇
10 - 50/ 50 (100)
That'll give you -40 / 50 (100)
Then,0.8 (100)
Answer: 80%

Then if it increased⬇
50 - 10 / 10 (100)
40 / 10 (100)
4(100)
400% would be your answer

The other reason⬇
10(50)
10 ÷ 50 = .1
.1(50)
Answer: 5
-----------------------------------------------
Then, again your answer could be : 55

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Which function represents a translation of the parent cubic function 8 units to the left? h(x) = x3 + 8 h(x) = x3 – 8 h(x) = (x

Ksju [112]

Using the definition of the Vertical shifts of graphs of the function :

"Suppose c>0,

To graph y=f(x)+c, shift the graph of y=f(x) upward c units.

To graph y=f(x)-c, shift the graph of y=f(x) downward c units"

Again we recall the definition of Horizontal shifts of graphs:

" suppose c>0,

the graph y=f(x-c), shift the graph of y=f(x) to the right by c units

the graph y=f(x+c), shift the graph of y=f(x) to the left by c units. "

consider $f(x)=x^3$ is the parent function.

$h(x)=x^3+8$ shifts the graph $f(x)=x^3$ upward by 8 units

$h(x)=x^3-8$ shifts the graph $f(x)=x^3$ downward by 8 units

$h(x)=(x+8)^3$ shifts the graph $f(x)=x^3$ left by 8 units

$h(x)=(x-8)^3$ shifts the graph $f(x)=x^3$ right by 8 units.

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2 years ago

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Which of the following is a true statement about the triangles shown on the graph?

boyakko [2]

Answer:

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2 years ago

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Answer:

When you cube root x to the power of 6, x becomes squared, and the exponent, 5, makes it x to the power of 10.

Step-by-step explanation:

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2 years ago

For a=4-5c when c=-1/2, what is the value of a? 1 is 1 half, not 1 divided by 2

Anika [276]

A = 4 - 5c
a = 4 - 5(-1/2)
a = 4 - (-2.5)
a = 4 + 2.5
a = 6.5

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2 years ago

Find the equation of the tangent line to the curve 2exy=x+y at (0,2).

DiKsa [7]

Answer:

The equation of the tangent line to the curve

3 x - y = 2

Step-by-step explanation:

Step(i):-

Given function = f(x,y) = $2e^{xy} -x-y=0$ ...(i)

Differentiating equation (i) with respective to 'x' , we get

$2 e^{xy} \frac{d}{d x} (x y) -1 -\frac{dy}{dx} =0$

apply formula

$\frac{d}{dx} (UV) = UV^{l} +V U^{l}$

step(ii):-

⇒ $2 e^{xy} (x(\frac{d}{d x} ( y))+y(1)) -1 -\frac{dy}{dx} =0$

⇒ $2 e^{xy} (x(\frac{d}{d x} ( y))+ 2e^{xy} y(1)) -1 -\frac{dy}{dx} =0$

Taking common d y/d x

$(2 e^{xy} (x) -1)\frac{dy}{dx} =1- 2e^{x y} y(1))$

$\frac{dy}{dx} =\frac{1- 2e^{x y} y(1))}{(2 e^{xy} (x) -1)}$

put At (0,2)

$\frac{dy}{dx} =\frac{1- 2e^{0} 2(1))}{(2 e^{0} (0) -1)}=\frac{1-4}{-1} =3$

slope of the curve m = 3

Step(iii):-

The equation of the tangent line to the curve

$y-y_{1} =m(x-x_{1} )$

y - 2 = 3 ( x - 0 )

3 x - y = 2

Final answer:-

The equation of the tangent line to the curve

3 x - y = 2

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2 years ago