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11 months ago

Pls help with math. :(

Mathematics

2 answers:

Karolina [17]11 months ago

8 0

Answer:

Tom will run 9 miles in 12 minutes.

Step-by-step explanation:

VMariaS [17]11 months ago

5 0

Answer:

I would say 9 miles

Step-by-step explanation:

He would have to be HAULING it to run 6 miles in 8 minutes lol.

But, I think it's 9 miles

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Does anyone know the answer to this one?

polet [3.4K]

Answer:

$\large\boxed{a(x+4)(x-5)=0,\ a\in\mathbb{R}-\{0\}}\\\\\text{for}\ a=1\to\boxed{x^2-x-20=0}$

Step-by-step explanation:

$\text{If}\ x_1\ \text{and}\ x_2\ \text{are the roots of the quadratic equation}\ ax^2+bx+c=0,\\\text{then}\ ax^2+bx+c=a(x-x_1)(x-x_2).\\\\\text{If}\ x_1\ \text{and}\ x_2\ \text{, then they are the roots of the quadratic equation.}\\\\\text{We have}\ x_1=-4\ \text{and}\ x_2=5.\ \text{Therefore we have the equation:}\\\\a(x-(-4))(x-5)=0\\\\a(x+4)(x-5)=0\qquad\text{for any value of}\ a\ \text{except 0}.$

7 0

3 years ago

2(4K + 7) Helppp lol

Hoochie [10]

8k+14

Hope this helps !!!

8 0

3 years ago

Read 2 more answers

Plz, HELP me with this question!

Tresset [83]

1/6 should be the right answer

7 0

3 years ago

100 POINTS!! OMG YOU NEED TO SOLVE THIS!

natita [175]

Answer:

The correct answer is: $\Delta BDE \cong \Delta BFK$ by rule ASA rule of congruence.

Step-by-step explanation:

First let us prove $\Delta BDE \cong \Delta BFK$ by rule ASA (rule of congruence).

Congruent side:

$\overline{BD} \cong \overline{BF}$ (Given)

Congruent angles:

1. By definition of perpendicular,

$\angle{BFK} = 90 \textdegree \ (Since \ \overline{FK} \ is \ perpendicular \ to \ \overline{AB} \ (\overline{FK} \perp \overline{AB} =Given))$

Also,

$\angle{BDE} = 90 \textdegree \ (Given)$

Therefore,

$\angle{BDE} \cong \angle{BFK}$

or you can say,

$\angle{D} \cong \angle{F}$

2. Common angle between $\Delta BDE \ and \ \Delta BFK$ is $\angle{B}$

In a nutshell, in $\Delta BDE$ ,

$\angle{B}$ (Angle)

$\overline{BD}$ (Side)

$\angle{BDE}$ (Angle)

are congruent to the following angle, side and angle of $\Delta BFK$ :

$\angle{B}$ (Angle)

$\overline{BF}$ (Side)

$\angle{BFK}$ (Angle)

Therefore, by ASA rule of congruence, we can say $\Delta BDE \cong \Delta BFK$ .

Since both triangles are congruent, the sides $\overline{ED}$ and $\overline{FK}$ are also congruent.

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3 years ago

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Need help with algerbra

TiliK225 [7]

Y^2+5y the greatest common factor is y so

y(y+5)

...

4x^2-49 this is a "difference of perfect squares" which always factors to:

(a^2-b^2)=(a-b)(a+b), in this case:

(2x-7)(2x+7)

...

5s^2-20 GCF is 5

5(s^2-4) now the parenthetic term is a difference of squares so

5(s-2)(s+2)

8 0

3 years ago