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1 year ago

Can you please HELP ME!! I have done question a but i need help on question b. For some people it may be easy but...

Mathematics

2 answers:

Nataliya [291]1 year ago

6 0

Step-by-step explanation:

a) 8x + 5x = 13x

b) 9f + 10f - 12f = 7f

hope it helps...

pickupchik [31]1 year ago

5 0

Answer:

1) 13x

2) 7f

Step-by-step explanation:

$1)8x + 5x \\ combine \: like \: terms \\ = 13x \\ \\ 2)9f + 10f - 12f \\ add \: 9f \: and \: 10f \: together = 19f \\ substract \: 19f \: from \: 12f \\ = 7f$

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the area of triangle ABC is 28 ft squared what is the area of the parallelogram ABCD 14 ft squared 56 ft squared 112 ft squared

Kazeer [188]

Answer:

Step-by-step explanation:

A∆ = BH/2

Area of a parm is BH (base x height) so we multiply area of triangle by 2.

28 x 2 = 56. There you go

6 0

2 years ago

Read 2 more answers

Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.

leonid [27]

<u>Answer-</u>

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

$12x-5y+14=0 \\ 12x-5y-14=0$

<u>Solution-</u>

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1$

$\Rightarrow 12h-5k-1=\pm 13$

$\Rightarrow 12h-5k=\pm 14$

$\Rightarrow 12h-5k=14,\ 12h-5k=-14$

$\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0$

$\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0$

Two equations are formed because one will be upper from the the given line and other will be below it.

4 0

2 years ago

Can you help me with this please???

Dahasolnce [82]

Answer:

Step-by-step explanation:

8 0

3 years ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

7 0

2 years ago

Read 2 more answers

Troy is writing a book of short stories. It is his

joja [24]

Answer:

91 short stories

Step-by-step explanation:

You just add 1 + 2 + 3 + 4 + 5 + 6... all the way to 13, and then at the end of 16 months, Troy still only wrote 91 stories since he only wrote them up until 13 months, and then stopped afterwards.

3 0

2 years ago