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3 years ago

HELP ME PLEASE THIS IS FOR CHORUS

Mathematics

1 answer:

KatRina [158]3 years ago

3 0

Answer:

1. (I'm not sure I'm sorry)

2. Ra

3. Si

4. Major Scale (my guess)

5. Chromatic Scale

6. enharmonic ; flat

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Find UV (Pic Stated Below)

natta225 [31]

Answer:

UV=29

Step-by-step explanation:

In right triangles AQB and AVB,

∠AQB = ∠AVB ...(i) {Right angles}

∠QBA = ∠VBA ...(ii) {Given that they are equal}

We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle

∠AQB+∠QBA+∠BAQ=180 ...(iii)

∠AVB+∠VBA+∠BAV=180 ...(iv)

using (iii) and (iv)

∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV

∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV (using (i) and (ii))

∠BAQ=∠BAV...(v)

Now consider triangles AQB and AVB;

∠BAQ=∠BAV {from (v)}

∠QBA = ∠VBA {from (ii)}

AB=AB {common side}

So using ASA, triangles AQB and AVB are congruent.

We know that corresponding sides of congruent triangles are equal.

Hence

AQ=AV

5x+9=7x+1

9-1=7x-5x

8=2x

divide both sides by 2

4=x

Now plug value of x=4 into UV=7x+1

UV=7*4+1=28+1=29

<u>Hence UV=29 is final answer.</u>

7 0

2 years ago

Need help with this please help

elena-14-01-66 [18.8K]

There is 24 avocados in each crate

Divide 8640 and 360 to get 24

Hope this will help you!

7 0

2 years ago

3<br>Find the value of a <br>in the equation,(3+4√3)(2-a√3)=-18+2√3

ankoles [38]

Given:

The equation is:

$\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}$

To find:

The value of a.

Solution:

We have,

$\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}$

On simplification, we get

$(3)(2)+(4\sqrt{3})(2)+(3)(-a\sqrt{3})+(4\sqrt{3})(-a\sqrt{3})=-18+2\sqrt{3}$

$6+8\sqrt{3}-3a\sqrt{3}-4a(3)=-18+2\sqrt{3}$

$6+8\sqrt{3}-3a\sqrt{3}-12a=-18+2\sqrt{3}$

$(6-12a)+(8-3a)\sqrt{3}=-18+2\sqrt{3}$

On comparing both sides, we get

$6-12a=-18$

$-12a=-18-6$

$a=\dfrac{-24}{-12}$

$a=2$

And,

$8-3a=2$

$-3a=2-8$

$a=\dfrac{-6}{-3}$

$a=2$

Therefore, the value of a is 2.

3 0

2 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

3 years ago

Someone pls solve this math puzzle.

Ivahew [28]

Answer:

x (blue circle thing) = 46

y (dna) = -63

z (molecules) = 52

Step-by-step explanation:

4 0

3 years ago