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shutvik [7]
3 years ago
6

8) Monica spent $34 on brownies and cookies. We know brownies cost $2

Mathematics
2 answers:
Viefleur [7K]3 years ago
7 0

Answer:

2b+.75c = $34

Step-by-step explanation:

b stands for the amount of brownies and the number beside it represents the price of the brownies. the c stands for the amount of cookies which is also unknown and the number beside that represents the price of the cookies. at the end of the equation is the total price of both the cookies and the brownies.

Anna [14]3 years ago
3 0
The answer is

2b + .75c = 34
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Suppose that about 58% of the people who are murdered actually knew the person who committed the murder. suppose that a detectiv
Fiesta28 [93]
Well, 100 minus 58 would be 42, so I'd say 42% is your answer.
5 0
3 years ago
Someone please solve this​
Wittaler [7]

Step-by-step explanation:

2 x  \times  \frac{1}{4}  - 1 \times  \frac{1}{4}  - 2 \frac{1}{4}  = 1 \times  \frac{1}{4}

\frac{2}{8}  -  \frac{1}{4}  -  \frac{2}{8}  =  \frac{1}{4}

and then minus them

8 0
2 years ago
2. What is the equation in slope-intercept form of the line that passes through thepoints (-4, 47) and (2, -16)?O21 979y=-2*+ 21
trapecia [35]

hello

the points given are (-4, 47) and (2, -16)

let's find the intercept of this equation

\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ x_2=2 \\ y_2=-16 \\ y_1=47 \\ x_1=-4 \end{gathered}\begin{gathered} m=\frac{-16-47}{2-(-4)} \\ m=-\frac{21}{2} \\ slope=-\frac{21}{2} \end{gathered}

now, since we know the value of the slope, we can use that in the standard equation on a straight line

the standard equation of a straight line is given as

\begin{gathered} y=mx+c \\ m=\text{slope} \\ c=\text{intercept} \end{gathered}

we can pick any of the points and solve for intercept

let's use (2, -16)

\begin{gathered} x=2 \\ y=-16 \\ y=mx+c \\ m=-\frac{21}{2} \\ -16=-\frac{21}{2}(2)+c \\ -16=-21+c \\ \text{collect like terms} \\ c=-16+21 \\ c=5 \end{gathered}

now we know the value of intercept (c) = 5 and the slope (m) = 21/2

let's use this to write equation of the straight line

\begin{gathered} y=mx+c \\ y=-\frac{21}{2}x+5 \end{gathered}

from the calculations above, the equation of the straight line is given as y = -21/2x + 5

8 0
1 year ago
The equation of the tangent plane to the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 at the point (x0, y0, z0) can be written as xx0 a2
svetlana [45]

Answer:

The equation of tangent plane to the hyperboloid

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=1.

Step-by-step explanation:

Given

The equation of ellipsoid

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1

The equation of tangent plane at the point \left(x_0,y_0,z_0\right)

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}=1  ( Given)

The equation of hyperboloid

\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1

F(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}[c^2}

F_x=\frac{2x}{a^2},F_y=\frac{2y}{b^2},F_z=-\frac{2z}{c^2}

(F_x,F_y,F_z)(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},-\frac{2z_0}{c^2}\right)

The equation of tangent plane at point \left(x_0,y_0,z_0\right)

\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)-\farc{2z_0}{c^2}(z-z_0)=0

The equation of tangent plane to the hyperboloid

\frac{2xx_0}{a^2}+\frac{2yy_0}{b^2}-\frac{2zz_0}{c^2}-2\left(\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}-\frac{z_0^2}{c^2}\right)=0

The equation of tangent plane

2\left(\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}\right)=2

Hence, the required equation of tangent plane to the hyperboloid

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}-\frac{zz_0}{c^2}=0

7 0
3 years ago
I need a answer right now I’ll give you points!
Kipish [7]

Answer:

<h2>G. 118⁰</h2>

100% right answer

8 0
3 years ago
Read 2 more answers
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