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3 years ago

•GIVEN : - 9a - 16 + 3 = 4 a •Solve for a •Show your work and check your work

Mathematics

1 answer:

jeka943 years ago

3 0

Answer:

-1

Step-by-step explanation:

-9a(-16+3)=4a ; solve for whole numbers first.

-9a-(13)=4a ; A negative 16 plus a positive 3 is 13.

-9a-4a=13 ; Switch 13 and 4a to make it easier to solve.

-13a=13 ; Now divide each side by -13.

a=-1

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Write each of the following expressions without using absolute value. |z−6|−|z−5|, if z<5

Andre45 [30]

Answer: 6 - 5

Step-by-step explanation:

|z - 6| - |z - 5| ; z < 5

Since z < 5, then

|z - 6| will be the absolute value of a negative number. Replace the absolute value with a negative and parentheses:

-(z - 6) = -z + 6

|z - 5| will be the absolute value of a negative number. Replace the absolute value with a negative and parentheses:

-(z - 5) = -z + 5

Now subtract them without the absolute value signs:

-z + 6 - (-z + 5)

Distribute the negative sign:

-z + 6 + z - 5

-z + z = 0 which leaves:

6 - 5

3 0

4 years ago

Read 2 more answers

The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your

Mashutka [201]

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

$D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Distance between two points P(-2,5) and Q(-1,1)

$PR = \sqrt{(-1+2)^{2}+(1-5)^{2} }$

$PR = \sqrt{17}$

Distance between two points Q(-1,1)and R(7,3)

$QR = \sqrt{(7+1)^{2} +(3-1)^{2} }$

$QR =\sqrt{68}$

Distance between two points R(7,3) and P(-2,5)

$RP =\sqrt{(-2-7)^{2} + (5-3)^{2} }$

$RP=\sqrt{85}$

now show that ∆PQR is a right triangle

$RP^{2} = PQ^{2} +QR^{2}$

Putting the value given above

$(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}$

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved

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4 years ago

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kakasveta [241]

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4 years ago

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Alex_Xolod [135]

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3 years ago

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LekaFEV [45]

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3 years ago