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3 years ago

14(a + 2) = 168 find a

Mathematics

1 answer:

valkas [14]3 years ago

8 0

$\bold\red{Answer:}$

14a+28=168

14a=140

a=10

hope it helps

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4 movie tickets cost $48. At this rate with is the cost of 5 movie tickets

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60 dollars, because each movie ticket cost 12$.

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3 years ago

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Y=-x^2+6x-4 Find the Axis of Symmetry and the Vertex, Also solve the whole equation.

Inessa [10]

Answer:

1. Axis of symmetry: x = 3

2. Vertex: (3,5)

3. Solution of the equation:

x-intercepts:

$(3-\sqrt{5},0) \\\\(3+\sqrt{5},0)$

y-intercept: (0, -4)

Explanation:

1. Equation:

$y=-x^2+6x-4$

2. Axis of symmetry:

That is the equation of a parabola, whose standard form is:

$y=ax^2+bx+c$

Where:

$a=-1;b=6;c=-4$

The axis of symmetry is the vertical line with equation:

$x=-b/2a$

Substitute $a=-1,\text{ and }b=6$

$x=-6/[(2)(-1)]=-6/(-2)=3$

Thus, the axis of symmetry is:

$x=3$

3. Vertex

The x-coordinate of the vertex is equal to the axys of symmetry, i.e x = 3.

To find the y-xoordinate, substitute this value of x into the equation for y:

$y=-x^2+6x-4\\\\y=-(3)^2+6(3)-4=-9+18-4=5$

Therefore, the vertex is (3, 5)

4. Find the x-intercepts

The x-intercepts are the roots of the equation, which are the points wher y = 0.

$y=-x^2+6x-4=0$

Use the quadratic equation:

$x=\frac{-b\pm \sqrt{b^2-4ac} }{2a} \\\\x=\frac{-6\pm\sqrt{(-6)^2-4(-1)(-4)} }{2(-1)}\\\\x_1=3-\sqrt{5} \\\\x_2=3+\sqrt{5}$

5. Find the y-intercept

The y-intercet is the value of y when x=0:

$y=-x^2+6x-4\\\\y=-(0)^2+6(0)-4=-4$

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3 years ago

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Answer: Both Liam and Tehya

Step-by-step explanation:

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Given that α and β are the roots of the quadratic equation <img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B6x-7%3Dp" id="Tex

siniylev [52]

Answer:

$\large \boxed{\sf \ \ \ p=-11 \ \ \ }$

Step-by-step explanation:

Hello,

$\alpha \text{ and } \beta \text{ are the roots of the following equation}$

$2x^2+6x-7=p$

It means that

$2\alpha^2+6\alpha-7=p \\\\2\beta ^2+6\beta -7=p \\\\$

And we know that

$\alpha= 2\cdot \beta$

So we got two equations

$2(2\beta)^2+6\cdot 2 \cdot \beta -7=p \\\\8\beta^2+12\beta -7=p\\\\ and \ 2\beta ^2+6\beta -7=p \ So \\\\\\8\beta^2+12\beta -7 = 2\beta ^2+6\beta -7\\\\6\beta^2+6\beta =0\\\\\beta(\beta+1)=0\\\\ \beta =0 \ or \ \beta=-1$

For $\beta =0, \ \ \alpha =0, \ \ p = -7$

For $\beta =-1, \ \ \alpha =-2, \ \ p= 2-6-7=-11, \ p=2*4-12-7=-11$

I assume that we are after two different roots so the solution for p is p=-11

b) $\alpha +2 =-2+2=0 \ and \ \beta+2=-1+2=1$

So a quadratic equation with the expected roots is

$x(x-1)=x^2-x$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago

What is 1,112 rounded to the nearest thousand?

DENIUS [597]

Answer: 1000

because you round down

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3 years ago