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3 years ago

Y=1000x+b how do i solve for b?

Mathematics

1 answer:

Irina18 [472]3 years ago

6 0

Answer:

put it into a graphing calculator, the point where the line intercepts the y-axis would be B

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Find the missing side. Round your answer to the nearest tenth.

Sergeu [11.5K]

Answer:

Step-by-step explanation:

ABC is a right angle triangle.

Therefore,

$cos \: (75) = \frac{ab}{ac}$

Plug the values

$cos \: (75) = \frac{x}{21}$

Apply cross product property

$x = 21 \times cos(75)$

Calculate

$x = 5.435$

After rounding to the nearest tenth, the answer will be:

$x = 5.4$

Hope this helps...

Best regards!!

6 0

4 years ago

Let $A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$. There exists a point $X$ and a constant $k$ such that for any point $P$,

goldfiish [28.3K]

Answer:

k=32

Step-by-step explanation:

Given the points:

$A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$.$

The first step is to find the <u>Centroid</u> of the triangle.

Centroid, X

$=\left(\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3} \right)\\=\left(\dfrac{4+6+(-1)}{3} ,\dfrac{-1+2+2}{3} \right)\\=\left(\dfrac{9}{3} ,\dfrac{3}{3} \right)=(3,1)$

Next, let P be a point (x,y)

Using the <u>distance formula, </u> $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ <u />

On Substitution into: $PA^2 + PB^2 + PC^2 = 3PX^2 + k$

$(x-4)^2+(y-(-1))^2+(x-6)^2+(y-2)^2+(x-(-1))^2+(y-2)^2=3[(x-3)^2+(y-1)^2]+k$

Let us simplify the LHS first

$\\LHS: x^2-8x+16+y^2+2y+1+x^2-12x+36+y^2\\-4y+4+x^2+2x+1+y^2-4y+4\\=3x^2-18x+3y^2-6y+62$

Also, the Right Hand Side

$RHS:3[(x-3)^2+(y-1)^2]+k\\=3[x^2-6x+9+y^2-2y+1]+k\\=3x^2-18x+27+3y^2-6y+3+k\\=3x^2+3y^2-18x-6y+30+k$

Therefore:

$3x^2-18x+3y^2-6y+62=3x^2+3y^2-18x-6y+30+k\\k=3x^2-18x+3y^2-6y+62-3x^2-3y^2+18x+6y-30\\k=3x^2-3x^2+3y^2-3y^2-18x+18x+62-30\\k=32$

7 0

3 years ago

Im gonna need help plz

guapka [62]

Answer:

its 4/64

if u want to simplfy its 1/16 :)

Step-by-step explanation:

i promise i did the math

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