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Margarita [4]
2 years ago
9

What is the solution set of the following equation? -8x + x + 15 = -7 x + 12 ​

Mathematics
1 answer:
UNO [17]2 years ago
3 0

Answer:

I dont k ow what the following set for your equation is...

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9.<br> The cost of 20 litres of petrol is £18<br> Work out the cost of 1 litre of petrol.
yulyashka [42]

Answer:

20 x 1 / 18= 1.11

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Which expression is equivalent to: (x +9)(x+3)
labwork [276]

Answer:

c

Step-by-step explanation:

x*x=x^2

x*9=9x

x*3=3x

9x+3x=12x

9x3=27

x^2+12x+27

6 0
3 years ago
Read 2 more answers
a man has 25 coins in his pocket all of which are dimes and quarters. if the total value of his change is 430 cents, how many di
valentinak56 [21]

Answer:16 quarters and 3 dimes

Step-by-step explanation:

430/25 (Quarters) = 17

25x16

(Because you can’t have 17 since 25x17=425 leaving no room for dimes.)

=400

Now 430-400=30

30 more cent is needed,

30/10(Dime value)

Is 3, so the answer is the man has 16 quarters and 3 dimes

7 0
3 years ago
Anyone mind helping me out?
Igoryamba

Answer:

611 square miles

Step-by-step explanation:

Area of triangle = (1/2)*a*b*sinC (formula when u are not given the height of the triangle, but u are given with the angle.)

= (1/2)*30*47*sin(60°)

= 610.54 square miles (2 d.p)

≈611 square miles (rounded to nearest mile)

:)

4 0
2 years ago
Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and
Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

\angle EAC = \angle ACD = 90^{\circ}

To prove :

EB\times BD=AB\times BC

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

\triangle AEC \sim \triangle EBA,

Similarly,

\triangle AEC \sim \triangle ABC

\implies\triangle EBA\sim \triangle ABC-----(1)

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

\triangle ADC \sim \triangle CBD,

Similarly,

\triangle ADC \sim \triangle ABC

\implies \triangle CBD\sim \triangle ABC-----(2)

From equations (1) and (2),

\triangle EBA\sim \triangle CBD

The corresponding sides of similar triangles are in same proportion,

\frac{EB}{BC}=\frac{AB}{BD}

\implies EB\times BD=AB\times BC

Hence, proved....

5 0
3 years ago
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