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2 years ago

Someone please help me my grades are bad

Mathematics

2 answers:

mamaluj [8]2 years ago

8 0

Answer:

AB=√3^2+3^2

AB=√18

AB=4.24

bija089 [108]2 years ago

5 0

Answer:

3√2 or 4.242641 units

Step-by-step explanation:

distance formula

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Start time 11:00 am Elapsed time 4 hr and 5 min

vivado [14]

Well, if it starts on 11:00 am, add 4 hours = 3:00 + 5 min = end time = 3:05 am.
Hopes this helps ;)

8 0

3 years ago

Read 2 more answers

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

tangare [24]

Answer:

$AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm$

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

$AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm$

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

$\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD$

Consider right triangle CAH.By the Pythagorean theorem,

$CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8$

The length cannot be negative, so CD=8 cm and

$CA=\sqrt{3}CD=8\sqrt{3}\ cm$

In right triangle ABC, angle B = 90° - 60° = 30°, leg AC is opposite to 30°, and the hypotenuse AB is twice the leg AC. Hence,

$AB=2CA=16\sqrt{3}\ cm$

By the Pythagorean theorem,

$BC^2=AB^2-AC^2\\ \\BC^2=(16\sqrt{3})^2-(8\sqrt{3})^2=256\cdot 3-64\cdot 3=576\\ \\BC=24\ cm$

3 0

2 years ago

Enter the degree of the polynomial below 7x7+10x4+4x3-5x11-10x6-6x7

natta225 [31]

When there is only one variable, the degree is the highest exponent.

Degrees of each term from L to R: 7, 4, 3, 11, 6, 7

The biggest of these is 11.

Final answer: 11

4 0

3 years ago

The general form of an parabola is 2x2−12x−3y+12=0 .

Murljashka [212]

Answer:

The standard form of the parabola is $(x-3)^2=4*\frac{3}{8}(y+2)$

Step-by-step explanation:

The standard form of a parabola is

$(x-h)^2=4p(y-k)$ .

In order to convert $2x^2-12x-3y+12=0$ into the standard form, we first separate the variables:

$2x^2-12x+3y+12=0\\\\2x^2-12x+12=3y$

we now divided both sides by 2 to remove the coefficient from $2x^2$ and get:

$x^2-6x+6=\frac{3}{2}y$ .

We complete the square on the left side by adding 3 to both sides:

$x^2-6x+6+3=\frac{3}{2}y+3$

$x^2-6x+9=\frac{3}{2}y+3$

$(x-3)^2=\frac{3}{2}y+3$

now we bring the right side into the form $4p(y-k)$ by first multiplying the equation by $\frac{2}{3}$ :

$\frac{2}{3} *(x-3)^2=\frac{2}{3} *(\frac{3}{2}y+3)\\\\\frac{2}{3} *(x-3)^2=y+2$

and then we multiplying both sides by $\frac{3}{2}$ to get

$(x-3)^2=\frac{3}{2} (y+2)$ .

Here we see that

$4p=\frac{3}{2}$

$\therefore p=\frac{3}{8}$

Thus, finally we have the equation of the parabola in the standard form:

$\boxed{(x-3)^2=4*\frac{3}{8}(y+2)}$

5 0

3 years ago

-6x+6y=6 -6x+3y=-12 How do you solve this problem?

sergij07 [2.7K]

Answer:

That's called system of equations. it is like adding and subtracting.

Step-by-step explanation:

first you want to find x or y. i will choose x. in order to do that you need to put one, ill do the first one but you can do first or second, and change it to negative (-6x+6y=6) or -(-6x+6y=6), changing it to 6x - 6y = -6. then the two x's will be cancled since 6-6 is 0. now you add 6y and 3y, and 6 with -12. get your answer, and do the same thing to y, too. sorry if its a bit confusing, took me a while to learn it too. this is highschool stuff so thats why you will although im im middle too!

3 0

3 years ago