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3 years ago

Help will give brainliest

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

8 0

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So the "back" of the arrow is on/above -20.

The arrow points to -12.

You added something to -20 to get to -12.

So the first number is -20. You added ? to get -12. -12 is the last number / the answer to the equation.

So the equation from what you have so far is -20 + ? = -12

You need to isolate the ? so that you know what it is.

So to get rid of the -20 (make it 0), you have to add 20. -20 plus 20 equals 0, which basically means it is gone. You also have to add 20 to -12. 20 plus -12 equals 8.

Now the equation is ? = 8.

So you know that the second number is 8.

The equation makes sense; -20 + 8 = -12.

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Expand (2 + a)9 using Pascal’s triangle.

Natalija [7]

Answer:

Step-by-step explanation:

The Pascal triangle is used to determine the coefficients of the terms when we expand the expression.

1 $(A + B) ^ 0$ = 1

1 1 $(A +B ) ^ 1 = 1A + 1B$

1 2 1 $(A+ B) ^ 2 = 1A^2 + 2 AB + 1B^2$

By extending the triangle, you will get the 9th row, which is your expression, of the coefficients. that is

1 9 36 84 126 126 84 36 9 1

Now, fill in AB in the gaps.

1AB + 9 AB + 36AB + 84AB + 126AB + 126AB +84AB + 36AB + 9AB + 1AB

Next, you need to go from the left to fill out the exponent of A and it will go down from 9 (the exponent of the whole thing) . That is

$1A^9B+9A^8B+36A^7B+84A^6B+126A^5B+126A^4B+84A^3B+36A^2B+9A^1B+1A^0B$

Next will be the exponent of B. this time, you go from the right and do the same with A. You can go from the left also, but go up from 0 to 9 for the exponent of B

$1A^9B^0+9A^8B^1+36A^7B^2+84A^6B^3+126A^5B^4+126A^4B^5+84A^3B^6+36A^2B^7+9A^1B^8+1A^0B^9$

The last step is just to simplify the A^0=1 and B^0 =1 at the first and the last terms.

$A^9+9A^8B^1+36A^7B^2+84A^6B^3+126A^5B^4+126A^4B^5+84A^3B^6+36A^2B^7+9A^1B^8+B^9$

Hope you can learn the method

5 0

3 years ago

What fractional part of 1 pound is an ounce?

bezimeni [28]

1 pound = 16 ounces
1 ounce =1/16 pound

3 0

4 years ago

jekas [21]

Given: In the given figure, there are two equilateral triangles having side 50 yards each and two sectors of radius (r) = 50 yards each with the sector angle θ = 120°

To Find: The length of the park's boundary to the nearest yard.

Calculation:

The length of the park's boundary (P) = 2× side of equilateral triangle + 2 × length of the arc

or, (P) = 2× 50 yards + 2× (2πr) ( θ ÷360°)

or, (P) = 2× 50 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)

or, (P) = 100 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)

or, (P) = 100 yards + 209.33 yards

or, (P) = 309.33 yards ≈309 yards

Hence, the option D:309 yards is the correct option.

7 0

4 years ago

Read 2 more answers

Help pls anyone? ty if u do :)

Schach [20]

Answer:

B. The division property of equality

Step-by-step explanation:

The division property of equality states that when we divide both sides of an equation by the same non-zero number, the two sides remain equal. That is, if a, b, and c are real numbers such that a = b and c ≠0, then a c = a c. Therefore B is the right answer.

pls give brainliest for the answer

4 0

2 years ago

Read 2 more answers

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago