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3 years ago

B is bewteen a and c if ab=4x+9 bc 5x+2 and ac= 56 find the value of ab

Mathematics

1 answer:

Ket [755]3 years ago

4 0

Answer:

It will be Ac because aB goes after Ac

Step-by-step explanation:

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Sergio [31]

1) The radius would be r = x + 4y

.... and the area would be A = pi ( x + 4y )^2 

2) x is not defined. You have to state the relationship of x with l ,w, and, h 

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7 0

3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Need ASAP

aniked [119]

Answer:

$x_1=\frac{1+\sqrt{33} }{2},y_1=\frac{5+\sqrt{33} }{2}\\ \\ x_2=\frac{1-\sqrt{33} }{2},y_2=\frac{5-\sqrt{33} }{2}$

Explanation:

You must find the solution of the system using the substitution method.

System:

$y=-x^2+2x+10\\\\ y=x+2$

Substitute x+2 for y:

$x+2=y=-x^2+2x+10$

Pass all the terms to one side:

$x^2-2x+x+2-10\\ \\ x^2-x-8=0$

Use the quadratic formula to solve:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\ \\ x=\frac{-(-1)+/-\sqrt{(-1)^2-4(1)(-8)} }{2(1)}\\ \\ x=\frac{1+/-\sqrt{1+32} }{2}$

$x_1=\frac{1+\sqrt{33} }{2}\\ \\x_2=\frac{1-\sqrt{33} }{2}$

Use y = x + 2 to find the value of y:

$y_1=2+x_1=2+\frac{1+\sqrt{33} }{2}\\ \\y_1=\frac{5+\sqrt{33} }{2} \\\\y_2=2+x_2=2+\frac{1-\sqrt{33} }{2}\\ \\ y_2=\frac{5-\sqrt{33} }{2}$

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3 years ago

Plz i need help on this

Degger [83]

Answer:

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3 years ago

What inequality is shown on the graph ?

Liula [17]

The inequality of the graph that is given is: y < 2x - 3.

<h3>How to Write the Inequality of a Graph?</h3>

First find the slope (m) = rise / run of the line.

Rise = 2 units

Run = 1 unit

Slope (m) = 2/1 = 2

Y-intercept (b) = -3.

Since the shaded side is at the left, we are going to use the ">" sign. Substitute m = 2 and b = -3 into y < mx + b.

y < 2x - 3

Learn more about inequality graph on:

brainly.com/question/371134

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7 0

2 years ago

Find the missing term. If y = 1 + i, then y3 − 3y2 + ( ? ) − 1 = -i .

attashe74 [19]

$y=1+i\qquad\qquad i^2=-1\\\\\\
y^3-3y^2+(?)-1 =-i\\\\(1+i)^3-3\cdot(1+i)^2+(?)-1+i=0\\\\1^3+3i+3i^2+i^3-3\cdot(1^2+2i+i^2)+(?)-1+i=0\\\\
1+3i+3\cdot(-1)+i\cdot i^2-3-6i-3i^2+(?)-1+i=0\\\\
1+3i-3+i\cdot(-1)-3-6i-3\cdot(-1)+(?)-1+i=0\\\\
1+3i-3-i-3-6i+3+(?)-1+i=0\\\\
-3-3i+(?)=0\\\\(?)=3+3i\\\\(?)=3(1+i)\\\\\boxed{(?)=3y}$

8 0

3 years ago

Read 2 more answers