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2 years ago

Help?????? Does anyone know the answer to this

Mathematics

1 answer:

Rama09 [41]2 years ago

5 0

I think it’s B, or 0.10x+0.15(50-x)=0.12(50)

Because 10% would give you 30ml, and 15% would equal 20ml to get 50ml, according to the equation.

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A medical company tested a new drug on 100 people for possible side

Harlamova29_29 [7]

Answer:

Step-by-step explanation:

As each person counts as 0.02 since theres 50 people in each, A has the correct conclusion and probability count of 0.12 (adult) and 0.40 (child)

6 0

2 years ago

the local high school has 19 honored students for every 81 non-honored students if there are 750 students in the school how many

aliina [53]

I got 176 I sad 750÷19 = 176

4 0

2 years ago

Does 5^9 x 6^7= 30^16 ?

Gemiola [76]

Yes. When solving a problem like this, multiply the base numbers normally, and add the exponents together.

If you need more help, comment below and I'd be happy to assist.

5 0

3 years ago

Graph the line that goes through the point (-6,-2) with slope of 1

vekshin1

Answer:

<h2>Check out the image.</h2>

Step-by-step explanation:

Let, the equation of the straight line is y = mx + c; where m is the slope of the equation and c is the constant.

The line has the slope 1.

Hence, the equation becomes y = x + c.

Since, the line passes through (-6, -2), -2 = -6 + c or, c = 4.

Hence, the equation of the straight line becomes y = x + 4.

Putting x = 0, we get, y = 4.

The line passes through (0, 4).

7 0

3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

2 years ago