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3 years ago

How many solutions does this system of equations have?

Mathematics

2 answers:

Andrew [12]3 years ago

8 0

I wanna say and go with D but im not quiet sure

Studentka2010 [4]3 years ago

6 0

C because y=5x-4 has only one solution to it compared to 5x+2 that has 2 or more

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Natalia goes to the bookstore with $25.00. She buys 4 magazines for $3.25 each and a comic book for $5.75. How much money does s

Andrew [12]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

How do u graph slope

Makovka662 [10]

You graph slope by going up and over based on what number you have as the slope. The starting point is the y intercept

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3 years ago

Can you explain to me (detailed) on how to do this problem? I will give you a brainly if you get this right.

7nadin3 [17]

Answer:

60 m^2

Step-by-step explanation:

To solve the area, solve the triangle and rectangle separately.

Solve the rectangle:

Multiply 10 and 4 to get the area of 40 m^2.

Solve the triangle:

Multiply 10 and 4, then divide by 2. The formula for a triangle's area is base*height/2. The area is 20.

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40+20 = 60 m^2

6 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:

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

6 0

3 years ago

Simplify 5+10−2^3 <br> thanks in advance

andrey2020 [161]

5+10-2^3
= 7

Step by step:
5+10-2^3
= 15 - 2^3
= 15 - 8
= 7

8 0

2 years ago