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vagabundo [1.1K]

2 years ago

9

PLS HELP!! Find the length of AB.

2 answers:

Georgia [21]2 years ago

8 0

Answer:

About 9.4 units

Step-by-step explanation:

chubhunter [2.5K]2 years ago

7 0

The answer is B

HOPE THIS IS HELPFUL

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What is the absolute deviation for 62 in the data set?

makvit [3.9K]

D is the answer for this question:))

8 0

3 years ago

PLs help 50 PTS!!!!! PLEASE ILL GIVE BRAINLIEST!!!!!

Nookie1986 [14]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2-x-4}$

Step-by-step explanation:

The equation of a parabola in vertex form:

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

<em>(h, k)</em><em> - vertex</em>

The focus is

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of <em>a</em> using $k+\dfrac{1}{4a}$

<em>k = -5</em>

$-5+\dfrac{1}{4a}=-4$ <em>add 5 to both sides</em>

$\dfrac{1}{4a}=1$ <em>multiply both sides by 4</em>

$4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4$

$\dfrac{1}{a}=4\to a=\dfrac{1}{4}$

Substitute

$a=\dfrac{1}{4},\ h=2,\ k=-5$

to the vertex form of an equation of a parabola:

$y=\dfrac{1}{4}(x-2)^2-5$

The standard form:

$y=ax^2+bx+c$

Convert using

$(a-b)^2=a^2-2ab+b^2$

$y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5$

<em>use the distributive property: a(b+c)=ab+ac</em>

$y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4$

3 0

3 years ago

Use a scientific calculator to find the logarithm for each number rounded ro four decimal places. Then state characteristic and

anzhelika [568]

Using a scientific calculator:
log 12 = 1.079181246
Rounded to four decimal places:
log 12 = 1.0792
Characteristic: 1; Mantis: 0.0792

Answer: OPtion B. 1.0792; 1; 0.0792

6 0

3 years ago

Help, quick! Worth 100 points

finlep [7]

Answer:

the answer is (C) 2(1-2x)=x-3

4 0

2 years ago

Find the remainder when x4 - 5x3 + x2 - 10x - 5 is divided by x + 3.

m_a_m_a [10]

Use the remainder theorem: we can decompose the given polynomial in terms of quotient and remainder polynomials $q(x)$ and $r(x)$ , respectively, such that

$x^4 - 5x^3 + x^2 - 10x - 5 = (x + 3) q(x) + r(x)$

Then letting x = -3 makes the quotient term vanish, and we're left with a remainder of

$r(-3) = (-3)^4 - 5\times(-3)^4 + (-3)^2 - 10\times(-3) - 5 = \boxed{250}$

3 0

2 years ago