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3 years ago

6

pls help Mary goes to dance class every 6 days. She goes to singing class every 8 days. When will Mary go for both dance and sin

ging classes on the same day?

1 answer:

kozerog [31]3 years ago

4 0

The third singing class

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Please help thank you

nikdorinn [45]

Answer:

See explanation

Step-by-step explanation:

$(x^2+9x+10)(x-2)= \\\\x^3-2x^2+9x^2-18x+10x-20= \\\\x^3+7x^2-8x-20$

Jimmy is incorrect.

Using long division, you can find that the real answer.

First, subtract $x^2(x-2)$ from the original expression, leaving you with $9x^2+2x-40$ .

Next, subtract $9x(x-2)$ from the expression, leaving you with $20x-40$ . Finally, subtract $20(x-2)$ from the expression, leaving you with a remainder of 0. This means that the real quotient is $x^2+9x+20$ . Hope this helps!

6 0

3 years ago

raketka [301]

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bit. $^{}$ ly/3a8Nt8n

5 0

3 years ago

(-3/4m)-(1/2)=2+(1/4,)

Aleks [24]

(-3/4m) -(1/2)=2+(1/4m)
-(1/2)=2+m
m= -3/2

5 0

3 years ago

Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y

I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

$\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }$

$\mathbf{ YX = \sqrt{(3)^2+(6)^2} }$

$\mathbf{ YX = \sqrt{9+36} }$

$\mathbf{ YX = \sqrt{45} }$

$\mathbf{ YX = \sqrt{9*5} }$

$\mathbf{ YX = 3 \sqrt{5} }$

Thus; the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

8 0

2 years ago

Write an equation in slope intercept form y-intercept 4 and slope is -3/5

dusya [7]

$\bf y=-\cfrac{3}{5}x+4\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

6 0

3 years ago