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Pie
2 years ago
5

Betty picks 4 random marbles from a bowl containing 3 white, 4 yellow, and 5 blue marbles.

Mathematics
1 answer:
mr Goodwill [35]2 years ago
4 0

Answer:

Step-by-step explanation:

hope it's helpful

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Daniel went to Target and bought a basketball for $ 21.85 and a shirt for $12.68. Daniel received $ 0.47 in change. How much did
jarptica [38.1K]

Answer:

$35

Step-by-step explanation:

To do this you would just add the 2 prices together and you would get $34.53 and you would add 47 cents to that and you would get that Daniel gave $35 to the cashier.

5 0
3 years ago
12/15 * 5/6
nydimaria [60]
<h2>¡Hola!</h2>

La multiplicación de las fracciones es 2/3

Para multiplicar las dos fracciones tenemos que multiplicar en línea: numerador por numerador y denominador por denominador.

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: {\sf \rightarrow  \dfrac{12}{15} \times  \dfrac{5}{6} }

Resolvamos:

\:  \:  \:  {\sf \rightarrow\dfrac{12}{15}\times \dfrac{5}{6}=\dfrac{12 \times 5}{15 \times 6}=\dfrac{60}{90} =\dfrac{2}{3} }

Por lo tanto, la multiplicación de las fracciones es 2/3

#Spj3

8 0
2 years ago
Read 2 more answers
Please help me find g(10).
timama [110]

Answer:

Step-by-step explanation:

g(10) = (10^2-8)/(10+4)= (100 - 8)/14 = 92/14 = 48/7 or 6 6/7

7 0
3 years ago
Plsss I need help!!!
mrs_skeptik [129]

Answer:

POSSIBLY  2

Step-by-step explanation: wasnt the best at math but i hope u do good

3 0
2 years ago
Use a matrix equation to solve the system of linear equations. left brace Start 2 By 1 Matrix 1st Row 1st Column 2nd Row 1st Col
natali 33 [55]

Answer:

\left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]

Step-by-step explanation:

The matrix system for the linear equations: x + 2y = 8, 2x + 6y = 9

\left[\begin{array}{ccc}1&2&\\2&6\\\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}8\\9\end{array}\right]

To get the coefficient of x and y, the inverse of the first matrix (let the first matrix be A) must be known.

A^{-1} = (1 / determinant of A) x Adjoint of A

the determinant of A = (1 x 6) - (2 x 2) = 6 - 4 = 2

Adjoint of A = \left[\begin{array}{ccc}6&-2&\\-2&1\\\end{array}\right]

A^{-1}= \frac{1}{2} \left[\begin{array}{ccc}6&-2\\-2&1\end{array}\right] = \left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]

4 0
2 years ago
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