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Lerok [7]
3 years ago
10

Help me please I can't get this

Mathematics
2 answers:
Vlad [161]3 years ago
8 0
8is the answer. Plug in 8 for x and solve
solong [7]3 years ago
3 0
If you put x=8 then you get y =8
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Which is equal to 42 • 48?<br> a. 416<br> b. 410<br> c. 46<br> d. 4−6
umka21 [38]
So With Questions Like These 
Remember, 
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5 0
4 years ago
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A theater group made appearances in two cities. The hotel charges before tax in the second city were $1500 lower than in the fir
KonstantinChe [14]

Answer:

The hotel bill in the first city before tax would be $ 650

Step-by-step explanation:

Let x be the hotel charges ( in dollars ) before tax in the first city,

∵  Hotel charges before tax in the second city were $1500 lower than in the first.

So, the hotel charges in second city = ( x - 1500 ) dollars,

Now, percentage of tax in,

First city = 8%

Second city = 10%,

So, the total tax in both city = 8% of x + 10% of (x-1500)

=\frac{8x}{100}+\frac{10(x-1500)}{100}

=0.8x + 0.10(x-1500)

According to the question,

0.8x + 0.10(x-1500)=435

0.8x + 0.10x - 150 = 435

0.9x = 435 + 150

0.9x = 585

\implies x = \frac{585}{0.9}=650

Hence, the hotel bill in the first city before tax would be $ 650.

7 0
3 years ago
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Solve for x.
Kay [80]

(4x+7)(2x+5) Like this???

4x+7=2x+5 or like this?!!?!

6 0
3 years ago
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As a ship approaches the dock, it forms a 70 angle between the dock and the lighthouse. At the lighthouse, an 80 angle is formed
Trava [24]

Answer:

The distance from the ship to the dock is approximately 5.24 miles

Step-by-step explanation:

From the parameters given in the question, we have;

The angle formed between the dock and the lighthouse = 70°

The angle formed between the dock and the lighthouse at the ship = 80°

The distance between dock and the lighthouse = 5 miles (From a similar question online)

By sine rule, we have;

\dfrac{a}{sin(A)} = \dfrac{b}{sin(B)} = \dfrac{c}{sin(C)}

Therefore, we have;

\dfrac{5}{sin(70^{\circ})} = \dfrac{The \ distance \ from \ the \ ship \ to \ the \ dock}{sin(80^{\circ})}

\therefore The \ distance \ from \ the \ ship \ to \ the \ dock = sin(80^{\circ}) \times \dfrac{5}{sin(70^{\circ})}

sin(80^{\circ}) \times \dfrac{5}{sin(70^{\circ})} \approx 5.24 \ mi

Therefore;

The distance from the ship to the dock ≈ 5.24 miles

7 0
3 years ago
Please help!!!!!
fgiga [73]

Answer:

  • 512/16= 32 miles per gallon
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