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lesantik [10]
3 years ago
9

Question 4 of 6, Step 3 of 4

Mathematics
1 answer:
Goshia [24]3 years ago
6 0
Well let’s see u don’t know the answer your question but I’m sure someone would love to help you
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Brianne can paint 120 square feet of walls in 34 hour.
kotegsom [21]
Good afternoon. 

120square --- 34h
 x   square ---   1h
      34x = 120
     x = 120/34
   x = 3,5 square
6 0
4 years ago
Plz help ................................
Elodia [21]

Answer:

1/4 is correct, or .25 in decimal form

Step-by-step explanation:

7 0
3 years ago
P(2,-8), Q(8,-8), R(8, -2), S(2,-2)<br> 180° rotation<br> I need help quickly
kompoz [17]

Answer:

P(-2,8)   Q(-8,8)   R(-8,2)  S(-2,2)

Step-by-step explanation: 180 degree rotation rule = (x,y) ---- (-x,-y)

ex : point B (-2, 7) will be B' (2, -7)

6 0
4 years ago
Θ lies in Quadrant II . sinθ=47sinθ=47 What is the exact value of cosθ in simplified form?
Pepsi [2]
I am assuming you mean 4/7 as the sine or cosine cannot be higher than 1.

Lets find <span>θ,
</span>
θ = [sin-1](4/7) = 34.85 °

But lets take into account that this value is the equivalent in Quadrant I.

If Θ lies in Quadrant II , then θ = 180 ° - 34.85 ° = 145.15 °

So cosθ = cos (145.15) = -0.821
4 0
3 years ago
A jeepney ride cost 11 pesos for the first 4 kilometers and each additional integer kilometers adds 3.50 pesos to the fare, use
Nookie1986 [14]

Answer:

f(d)=\left\{ \begin{matrix} 11 & \text{for} & 0< d\leq 4  \\-3+3.5d & \text{for}&  d > 4  \end

Step-by-step explanation:

Cost of first 4 kilometers ride = 11 pesos.

So, for the first 4 km, the fair is constant i.e.

for 1 km the fare is 11 pesos,

for 2 km the fair is also 11 pesos,

similarly, for 3 km of 4 km, the fare is 11 pesos.

Hence, for the distance 0\geq d\geq 4, the fair function,

f(d)=11\cdots(i)

After 4 km, there is an increment of $ 3.50 for each kilometer.

So, the fare function up to 5 kilometers,

f(d)=11+3.5=14.5

So, the fare function up to 6 kilometers, i.e for the distance 5,

f(d)=11+2\times3.5=18

This can be arranged as the fare function up to 6 km,

f(d)=11+(6-4)\times3.5=18

Similarly, the fare function up to d kilometer (n>4),

f(d)=11+(d-4)\times3.5

\Rightarrrow f(d)=-3+3.5d\cdots(ii)

Hence, from equations (i) and (ii),

f(d)=\left\{ \begin{matrix} 11 & \text{for}\; 0< d \leq 4  \\-3+3.5d & \text{for}\;  d > 4  \end

8 0
3 years ago
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