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3 years ago

9

since this was my first time you guys are in for a treat .... FREE POINTS answer question " 1+0 " don't for get to friend me :)

see y'all next week

2 answers:

ollegr [7]3 years ago

6 0

Answer:

1

Step-by-step explanation:

1+0=1

irina1246 [14]3 years ago

3 0

Hii 1+0=1 sorry don’t know to friend:(

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Use the graph below for this question:

Anastaziya [24]

Answer:

-4

Step-by-step explanation:

Do (y1-y2)/(x1-x2) so (3-(-5))/(0-2)

8/-2=-4

8 0

2 years ago

The weight in pounds of a baby in the first six months of life can be modeled by the

gavmur [86]

Answer:

14.5

Step-by-step explanation:

You would plug 5 into the equation for m and have w = 1.5 (5) + 7

Then you would multiply 1.5 and 5 and get 7.5 and then add 7 and get 14.5

6 0

3 years ago

A car traveled for 6 hours on one highway and for 2 hours on a second highway. If the total distance traveled was 488 miles, and

Nataly [62]

6V + 2(V+12) = 488
8V = 464
V = 58 mph for 6 hours

V+12=70 mph for 2 hours.

check: 58*6 = 348

70*2 = 140

348 +140 =488

the car was traveling at 70 mph on the 2nd highway

7 0

3 years ago

If g(x)=3/4x+2,find g(-12)

Triss [41]

Answer:

g(-12) = -7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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