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3 years ago

10

Evaluate the equation. X=log27(3)

1 answer:

borishaifa [10]3 years ago

6 0

From what i interpreted this question as... this is what i got and i hope i helped a bit

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9) The present ages of Juna and Jubin are in the ratio 5:8. After 4 years the sum

vodka [1.7K]

Answer:

Present age of Juna = 10 years

Present age of Jubin = 16 years

Step-by-step explanation:

Ratio of present ages = 5: 8

Let the present ages of:

Juna be 5x years and Jubin be 8x years

(where x is the common multiplier of the given ratios)

Four years after:

Juna's age = (5x + 4) years

Jubin's age = (8x + 4) years

According to the question:

(5x + 4) + (8x + 4) = 34

13x + 8 = 34

13x = 34 - 8

13x = 26

x = 26/13

x = 2

5x = 5*2 = 10 years

8x = 8*2 = 16 years

Hence,

Present age of Juna = 10 years

Present age of Jubin = 16 years

4 0

3 years ago

3. Farhana goes to McDonald's with her friends. They buy 2 Big

hodyreva [135]

25 is your answer , thanks

4 0

3 years ago

Reference angle to 0=225

Nuetrik [128]

Since the angle 225° is in the third quadrant, the reference angle formula is Ar=Ac-180°. Ar = 225° − 180°

Ar=225°-180°

The reference angle is Ar=45°

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7 0

4 years ago

Read 2 more answers

Pls help asap, I put attachments down below, I will also give brainliest

lesya [120]

Answer:

Question 1;

∴ The y values are 1.49 , 2.98 , 4.47 and 5.96

Question 2;

∴ The y values are 2.00 , 2.8 , 3.6 and 4.4

Question 3;

∴ The y values are -9 , -3 , 3 and 9

Question 4;

∴ y values are $\frac{1}{2}$ , $\frac{3}{4}$ , 1 and $\frac{5}{4}$

Step-by-step explanation:

Question 1;

Given data,

Apples cost $\$1.49$ per pound.

$x$ = weight in pound

$y$ = Cost in $\$$

Here the relation between $x$ and $y$ comes;

$y=x\times 1.49$ (equation 1)

Let, Plug all x value in equation 1;

when $x=1$ then $y=1\times 1.49=1.49$ ,

when $x=2$ then $y=2\times 1.49=2.98$ ,

when $x=3$ then $y=3\times 1.49=4.47$ ,

when $x=4$ then $y=4\times 1.49=5.96$ ,

∴ The y values are 1.49 , 2.98 , 4.47 and 5.96

Question 2;

From question,

Initial fees of cab is $\$$ 2.00 and Passengers are charged $\$$ 0.80 for every minute.

Let,

$x$ = Time in minutes

$y$ = Amount in $\$$

Here the relation between x and y comes;

$y=x\times 0.80 +2.00$ (equation 2)

Let,By plug all x value in equation 2;

When $x=0$ then $y=0\times 0.80+2.00=2.00$ ,

When $x=1$ then $y=1\times 0.80+2.00=2.8$ ,

When $x=2$ then $y=2\times 0.80+2.00=3.6$ ,

When $x=3$ then $y=3\times 0.80+2.00=4.4$ ,

∴ The y values are 2.00 , 2.8 , 3.6 and 4.4

Question 3;

Given relation between $x$ and $y$ ;

$y=6\times x +3$

Plug all the values of x in above equation,

When $x=-2$ then $y=(6\times -2) +3=-9$ ,

When $x=-1$ then $y=(6\times --1) +3=-3$ ,

When $x=0$ then $y=(6\times 0) +3=3$ ,

When $x=1$ then $y=(6\times 1) +3=9$ ,

∴ The y values are -9 , -3 , 3 and 9

Question 4;

Given relation between $x$ and $y$ ;

$y=(\frac{1}{4} \times x)+1$

Plug all the values of x in above equation,

When $x=-2$ then $y=(\frac{1}{4} \times -2)+1=\frac{1}{2}$ ,

When $x=-1$ then $y=(\frac{1}{4} \times -1)+1=\frac{3}{4}$ ,

When $x=0$ then $y=(\frac{1}{4} \times 0)+1=1$ ,

When $x=1$ then $y=(\frac{1}{4} \times 1)+1=\frac{5}{4}$ ,

∴ y values are $\frac{1}{2}$ , $\frac{3}{4}$ , 1 and $\frac{5}{4}$

7 0

4 years ago

W = 13, x = 9, y = 1.9, z = 0.5<br><br><br> x2 + 4

prisoha [69]

Answer:

17 is the answer

Step-by-step explanation:

6 0

3 years ago