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3 years ago

Can someone help me please?-

Mathematics

2 answers:

stiv31 [10]3 years ago

6 0

The answer is D. Solve for the first variable in one of the equations, then substitute the result into the other equation.

Oliga [24]3 years ago

4 0

The answer is d. Have a good day

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Write a paragraph proof.

worty [1.4K]

It is given that line segment BC is congruent to line segment EC and that line segment AC is congruent to DC. Because of the vertical angles theorem, angle BCA is equal to angle DCE. Therefore, triangles CBA AND DEC are congruent by SAS. Using CPCTC, BA is equal to ED.

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3 years ago

Need help figuring out the values of x and y for 11-13 please and thank you!!

Ne4ueva [31]

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12 is 90 degrees for x and y

13 is also 90 degrees for x and y

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3 years ago

Area of a rectangle is 36 what is x if the width is x and length is x+5

Reika [66]

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5 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

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3 years ago

The Empire State Building weights about 7.3 x 10 ^8

kvasek [131]

Answer:

730000000 lbs?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers