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nekit [7.7K]
2 years ago
15

His grandmother is sick, so Hassan is making a pot of bean soup to bring for dinner. The recipe calls for equal amounts of 3 kin

ds of beans: kidney, pinto, and black. In all, Hassan uses 2 1/4 cups of beans in the soup.
Use an equation to find the amount of each kind of bean Hassan uses in the soup.
Mathematics
1 answer:
lions [1.4K]2 years ago
7 0

Answer:

0.75 cup

Step-by-step explanation:

K + p + b = 2 1/4

K = p = b

Sence all beans are equal we can divide 2 1/4 by 3

2 1/4 / 3 = 3/4

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4y − 3x − 2y 11 x − 2 x
Inessa05 [86]
By rearranging it:
-3x-2x-22xy+4y
=-5x-22xy+4y answer
4 0
3 years ago
Question Help
lara [203]

Answer:

4324eqW

Step-by-step explanation:

3 + 5 =7

7 0
3 years ago
The intensity of a light source at a distance is directly proportional to the strength of the source and inversely proportional
jolli1 [7]

Answer:

x=\frac{16}{\sqrt[3]{2}+1 }

Step-by-step explanation:

Q= illumination

I = intensity

Q= I/d^2

Q_total = \frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}

= \frac{I}{x^2}+\frac{2I}{(16-x)^2}

now Q' = 0

⇒I{-\frac{2}{x^3}}+\frac{4}{(16-x)^3}

x=\frac{16}{\sqrt[3]{2}+1 }

[/tex]\frac{1}{x^3} = \frac{2}{(16-x)^3}

x=\frac{16}{\sqrt[3]{2}+1 }

is the required point

5 0
3 years ago
In ΔVWX, the measure of ∠X=90°, WX = 8.3 feet, and XV = 2.5 feet. Find the measure of ∠W to the nearest tenth of a degree.
kiruha [24]

Answer:

16.76°

Step-by-step explanation:

In ΔVWX, the measure of ∠X=90°, WX = 8.3 feet, and XV = 2.5 feet.

We want to find the measure of <W.

We know side length that is adjacent and opposite to <W.

We can use the tangent ratio, to find the measure of <W.

The tangent ratio is opposite over hypotenuse.

\tan(m \angle \: w)  =  \frac{2.5}{8.3}

\tan(m \angle \: w)  =  0.301

Take tangent inverse to get:

m \angle \: w= { \tan}^{ - 1}   (0.301)

m \angle \: w=16.76  \degree

7 0
3 years ago
Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
Anit [1.1K]

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

4 0
3 years ago
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