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3 years ago

Need help w this due today

Mathematics

1 answer:

8_murik_8 [283]3 years ago

5 0

Answer:

Step-by-step explanation:

you have a right triangle,

a^2+b^2=c^2

10^2=4^2+b^2

100=16+b^2

100-16=b^2

84=b^2

b=9.16

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15/3 = 5 4_7 + 5 2_3<br> b=?

OLEGan [10]

Answer:

válla si son brutos está bien fácil esa pregunta lo que pasa es que estoy haciendo mi trabajo

5 0

3 years ago

Read 2 more answers

A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the

makkiz [27]

$f(t)=h=-16t^2+20t+6

$

a=-16, b=20, c =6

Δ= $b^2-4ac=20^2-4*(-16)*6=400+384=784$

$\sqrt{Δ}=\sqrt{784}=28$

max:

$p=-\frac{b}{2a}=-\frac{20}{-32}=\frac{20}{32}=\frac{5}{8}$

max heigh: $h=-16(\frac{5}{8})^2+20*\frac{5}{8}+6=-16\frac{25}{64}+\frac{100}{8}+6$

$h=-\frac{25}{4}+\frac{50}{4}+6=\frac{-25+50+24}{4}=\frac{49}{4}$

$h=\frac{49}{4}$ at $t=\frac{5}{8}s$

will fall down at t:

$t=\frac{-20-28}{-32}=\frac{-48}{-32}=\frac{3}{2}=1.5$

$t=1.5s$

8 0

3 years ago

Between what two integers does 98 fall between

Nat2105 [25]

Answer: 99 and 97

Step-by-step explanation: If you make a numberline and write the numbers 90 through 100. 98 is in between integers 99 and 97.

8 0

3 years ago

If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c

dolphi86 [110]

Given:

$(ax+2)(bx+7)=15x^2+cx+14$

And

$a+b=8$

Required:

To find the two possible values of c.

Explanation:

Consider

$\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}$

$\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}$

And also given

$a+b=8---(2)$

Now from (1) and (2), we get

$\begin{gathered} a+\frac{15}{a}=8 \\ \\ a^2+15=8a \\ \\ a^2-8a+15=0 \end{gathered}$ $a=3,5$

Now put a in (1) we get

$\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}$

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

$\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}$

When a=5 and b=3, we have

$\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}$

Final Answer:

The option D is correct.

31 and 41

8 0

1 year ago

I went to the candy store last weekend and I bought a total of 10 packs of candy. I bought M&Ms for $1 per pack and Twizzler

mario62 [17]

Let's see. We can clarify a few things to make the question easier.

-We bought 10 packs of candy
-M&Ms are $1 per pack
-Twizzlers are $0.5 per pack

We need to find out how many bags of M&Ms we bought by only spending $6.

There are many combinations you can use to get to $6. However, I'm not sure what the format is. For example, you could buy 4 twizzler packs and then use the remaining money to find out how many M&Ms you bought. But this seems to be open to answers. Sorry

8 0

3 years ago