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olganol [36]
3 years ago
7

Write the following expression in standard place-value form.

Mathematics
1 answer:
erastova [34]3 years ago
4 0

Answer:

3408

Step-by-step explanation:

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Complete the dilation centered at A with a scale factor of 1/2.
ElenaW [278]

Answer:

A'=(5/2,-5/2)

therefore A'(2.5,-2.5)

Step-by-step explanation:

8 0
3 years ago
In MON, J, K, and L are midpoints. If JL = 11, LK = 13, and ON = 20, and JL || MN, LK || MO, and JK || ON, what is the length of
In-s [12.5K]

Answer:

The lengths of MN is 22 units, MO is 26 units and JK is 10 units

Step-by-step explanation:

<em>A l</em><em>ine segment</em><em> joining the </em><em>mid-points of two sides</em><em> in a triangle is </em><em>parallel to the third side</em><em> and </em><em>equal to half its length</em>

In Δ MON

∵ J, K, and L are mid-points

∵ JL // MN and LK // MO

∴ L is the mid-point of ON

∴ J is the mid-point of MO

∴ K is the mid-point of MN

∵ J, L are the mid-points of MO and ON

∵ JL is opposite to MN

→ By using the rule above

∴ JL = \frac{1}{2} MN

∵ JL = 11 units

∴ 11 = \frac{1}{2} MN

→ Multiply both sides by 2

∴ 22 = MN

∴ MN = 22 units

∵ K, L are the mid-points of MN and ON

∵ KL is opposite to MO

→ By using the rule above

∴ KL = \frac{1}{2} MO

∵ KL = 13 units

∴ 13 = \frac{1}{2} MO

→ Multiply both sides by 2

∴ 26 = MO

∴ MO = 26 units

∵ J, K are the mid-points of MO and MN

∵ JK is opposite to ON

→ By using the rule above

∴ JK = \frac{1}{2} ON

∵ ON =20 units

∴ JK = \frac{1}{2} (20)

∴ JK = 10 units

∴ The lengths of MN are 22 units, MO is 26 units and JK is 10 units

3 0
3 years ago
The range for the given domain of the function f(x)=3x-6/x+4.5 is
mash [69]
Hello,

We can not divide by 0.

==> x+4.5≠0
==>x≠-4.5

dom f=R\{4.5}
4 0
3 years ago
Solve for a in the figure shown.
aleksandrvk [35]

Answer: 58 is the answer

hope this help and here my steps use a mark or a pin

and line it with the line if it's wrong sorry for the wrong

answer

7 0
3 years ago
Read 2 more answers
How do you graph a possible situation?
Svet_ta [14]
To graph a situation that would involve a linear graph, first determine your x and y axes.
The x-axis will be the independent variable, one that does not change based on other variables. An example is time.

The y-axis, the dependent variable, depends on the independent variable.

The model equation for a linear line is y = mx + b.
"m" is the slope, and the "b" is the y-intercept (where the graph crosses the x-axis at x=0).

For example, a situtation could be that Joe starts with $10 in his account and adds $5 every day to his account.
The x-axis is time in days.
The y-axis is amount of money in his account.
The slope, or rate of change is 5.
The y-intercept, the amount of money he has at x=0 (0 days) is $10.

The equation would be y = 5x + 10
To draw this, plot the y-intercept at (0, 10), and the next point would be 5 units up and one unit to the right because the slope is 5, or 5/1 (remember slope is rise over run: "rise" up 5 and "over" to the right 1).

4 0
3 years ago
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