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2 years ago

Dylan weighs 45 pounds. Use the expression you derived in part I to calculate how much more Bruce weighs than Alan.

Mathematics

1 answer:

boyakko [2]2 years ago

3 0

Hi I’m sorry Alan I got home

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Which figure can be formed from the net?

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I dont know if its 100% correct but I believe its this one

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2 years ago

Juniors and seniors at a local school were surveyed to see if they were left-handed or right-handed. The data is displayed in th

Novosadov [1.4K]

P(left handed senior) = 14/(36 + 14 + 44 + 106) = 14/200 = 0.07 = 7%

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3 years ago

Find tan A for the triangle below.<br><br> a 1.5<br> b. 0.667<br> C. 0.832<br> d. 1.202

Kaylis [27]

Answer:

Step-by-step explanation:

on edge

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2 years ago

Solve for x {x+4]= 6

wlad13 [49]

The solution to the equation x(x+4) = 6 is x = -2 + √10 or x = -2 - √10 after solving with the quadratic formula.

<h3>What is a quadratic equation?</h3>

Any equation of the form $\rm ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

$\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$

We have an equation:

x(x + 4) = 6

By distributive property:

x² + 4x = 6

x² + 4x - 6 = 0

a = 1, b = 4, c = -6

Plugging all the values in the formula:

$\rm x = \dfrac{-4 \pm\sqrt{4^2-4(1)(-6)}}{2(1)}$

After calculating:

$\rm x = \dfrac{-4 \pm\sqrt{40}}{2}$

$\rm x=-2+\sqrt{10},\ \ \ or \ \ \ x=-2-\sqrt{10}$

Thus, the solution to the equation x(x+4) = 6 is x = -2 + √10 or x = -2 - √10 after solving with the quadratic formula.

Learn more about quadratic equations here:

brainly.com/question/2263981

#SPJ1

6 0

2 years ago

Read 2 more answers

Explain how to find the vertex of 2x^2+12x+16

Elina [12.6K]

The vertex form:

$y=a(x-h)^2+k\\\\(h,\ k)-vertex$

Use $(a+b)^2=a^2+2ab+b^2\qquad(*)$

$2x^2+12x+16=2(x^2+6x)+16=2(x^2+2(x)(3))+16\\\\=2(\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2)+16=2((x+3)^2-9)+16\\\\=2(x+3)^2+(2)(-9)+16=2(x+3)^2-18+16\\\\=\boxed{2(x-(-3))^2-2}\to h=-3,\ k=-2$

<h3>Answer: (-3, -2).</h3>

Other method:

$f(x)=ax^2+bx+c\\\\vertex:(h,\ k)\\\\h=\dfrac{-b}{2a},\ k=f(h)$

We have

$f(x)=2x^2+12x+16\\\\a=2,\ b=12,\ c=16$

Substitute:

$h=\dfrac{-12}{(2)(2)}=\dfrac{-12}{4}=-3\\\\k=f(-3)=2(-3)^2+12(-3)+16=2(9)-36+16=18-36+16=-2$

<h3>Answer: (-3, -2).</h3>

5 0

3 years ago