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Over [174]
4 years ago
13

Write two subtraction equations to show how to find 15 - 7

Mathematics
1 answer:
alex41 [277]4 years ago
3 0
These are the two equation

10-2
13-5
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In 14 days the daily high temperature in cedar hills dropped by 16.8 degrees Fahrenheit what is the average daily rate of change
Montano1993 [528]

Answer:

1.2°F

Step-by-step explanation:

Given that:

The drop in temperature in degree Fahrenheit over a period of 14 days = 16.8

The average daily rate of change in degwrr Fahrenheit ;

Total drop in temperature / number of days

= 16.8°F / 14

= 1.2°F

Hence, the average daily drop in temperature is 1.2°F

6 0
3 years ago
What kind of change occurs when a figure is translated?
vagabundo [1.1K]
It depends on what way the figure is translated, but it's possible it could be a dilations, reflection, rotation, etc.
3 0
3 years ago
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Please Help!! Which graph shows the solution to this system of inequalities?
scoundrel [369]

Answer:

A. The first graph

Step-by-step explanation:

When you graph these two equations on a graphing calc, you should be able to see one dotted line and one solid, as well as the solutions being mostly in the 1st quadrant.

7 0
4 years ago
Pls pls pls pls helpppppp
pychu [463]

Answer:

$32.25

Step-by-step explanation:

$53.75 ÷ 5 = $10.75

$10.75 × 3 = $32.25

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3 years ago
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In the trapezoid ABCD (AB∥CD) point M∈AD, so that AM:MD=3:5. Line L ∥AB and going through point M intersects diagonal AC and leg
siniylev [52]

Answer:

\dfrac{AP}{PC}=\dfrac{3}{5}

\dfrac{BN}{CN}=\dfrac{3}{5}

Step-by-step explanation:

Consider triangles AMP and ADC. In these triangles,

  • angle A is the common angle, so \angle MAP\cong \angle DAC by reflexive property;
  • angles AMP and ADC are congruent as corresponding angles when two parallel lines MP and CD are cut by transversal AD.

Hence, triangles AMP and ADC are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

\dfrac{AM}{AD}=\dfrac{AP}{AC}\\ \\\dfrac{3x}{3x+5x}=\dfrac{AP}{AC}\\ \\\dfrac{AP}{AC}=\dfrac{3}{8}\Rightarrow AP=\dfrac{3}{8}AC\\ \\PC=AC-AP=AC-\dfrac{3}{8}AC=\dfrac{5}{8}AC,

so

\dfrac{AP}{PC}=\dfrac{\frac{3}{8}AC}{\frac{5}{8}AC}=\dfrac{3}{5}

Consider triangles ACB and PCN. In these triangles,

  • angle C is the common angle, so \angle ACB\cong \angle PCN by reflexive property;
  • angles ABC and PCN are congruent as corresponding angles when two parallel lines PN and AB are cut by transversal BC.

Hence, triangles ACB and PCN are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

\dfrac{CP}{AP}=\dfrac{CN}{CB}\\ \\\dfrac{5x}{3x+5x}=\dfrac{CN}{CB}\\ \\\dfrac{CN}{CB}=\dfrac{5}{8}\Rightarrow CN=\dfrac{5}{8}CB\\ \\BN=BC-CN=BC-\dfrac{5}{8}BC=\dfrac{3}{8}BC,

so

\dfrac{BN}{CN}=\dfrac{\frac{3}{8}BC}{\frac{5}{8}BC}=\dfrac{3}{5}

4 0
3 years ago
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