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evablogger [386]

1 year ago

15

Bob Ross bought a Blackberry on the Web for $90.49. He saw the same Blackberry in the mall for $120.21. How much did Bob save by

buying on the Web?
Note: Round your answer to the nearest cent.

1 answer:

liraira [26]1 year ago

5 0

Answer:

$29.72

Step-by-step explanation:

$120.21-$90.49=$29.72

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2 pounds 13 ounces = how many ounces? *<br> Please help i will give brainliest :P

wolverine [178]

Answer:

45 ounces

Step-by-step explanation:

8 0

2 years ago

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blondinia [14]

Answer: $\displaystyle\ \large \boldsymbol{ 2t^2+5t+2= {(2t+1) (t+2)}}$

Step-by-step explanation:

$\displaystyle\ \large \boldsymbol{} 2t^2+5t+2=0 \\\\2t^2+\underbrace{4t+t}_{5t}+2=0 \\\\ (2t^2+4t)+(t+2)=0 \\\\ 2t(\underline{t+2})+(\underline{t+2})=0 \\\\ \boxed{(2t+1) (t+2)=0 }$

3 0

2 years ago

After a dilation with a center of (0, 0), a point was mapped as (4, –6) → (12, y). A student determined y to be –2. Evaluate the

SpyIntel [72]

Answer:

B. The student incorrectly calculated the scale factor to be –2

Step-by-step explanation:

Given that :

After a dilation with a center of (0, 0), a point was mapped as (4, –6) → (12, y).

The student determined y to be -2

If a figure dilated with a center of (0, 0) and scale factor k, then

(x , y) → (kx , ky)

(4, -6) → (12, y)

$k = \dfrac{x'}{x}$

$k = \dfrac{12}{4}$

k = 3

Thus; the scale factor is 3

Now; the y-coordinate can now be calculated as;

ky = (3 × -6)

ky = -18

Therefore; the value of y = -18 and the student incorrectly calculated the scale factor to be -2.

3 0

2 years ago

Write each set of fractions with the lowest common denominator and then find which fraction is greater. 4/7 and 5/8 3/8 and 4/9

Elan Coil [88]

Answer:

$\frac{32}{56}\ < \frac{35}{56}$

$\frac{27}{72} \ < \frac{32}{72}$

$\frac{20}{24}\ < \frac{21}{24}$

$\frac{9}{30}\ >\frac{8}{30}$

Step-by-step explanation:

Given that,

First set of fraction is $\frac{4}{7}\ and \ \frac{5}{8}$ .

Second set of fraction is $\frac{3}{8}\ and \ \frac{4}{9}$ .

Third set of fraction is $\frac{5}{6}\ and \ \frac{7}{8}$ .

Fourth set of fraction is $\frac{3}{10}\ and \ \frac{4}{15}$ .

Now,

Considering the first set of fraction:

LCM of 7 and 8 is 56. Now, multiply by 8 to both numerator and denominator to $\frac{4}{7}$ . and multiply by 7 to both the numerator and denominator to $\frac{5}{8}$ . We get new set of fraction as $\frac{32}{56}\ and\ \frac{35}{56}$ .

∴ $\frac{32}{56}\ < \frac{35}{56}$

Again considering the second set of fraction:

LCM of 8 and 9 is 72. Now, multiply by 9 to both the numerator and denominator to $\frac{3}{8}$ . and multiply by 8 to both the numerator and denominator to $\frac{4}{9}$ . We get new set of fraction as $\frac{27}{72} \ and \ \frac{32}{72}$ .

∴ $\frac{27}{72} \ < \frac{32}{72}$ .

Again considering the third set of fraction:

LCM of 6 and 8 is 24. Now, multiply by 4 to both the numerator and denominator to $\frac{5}{6}$ . and multiply by 3 to both the numerator and denominator to $\frac{7}{8}$ . We get the new set of fraction as $\frac{20}{24}\ and \ \frac{21}{24}$ .

∴ $\frac{20}{24}\ < \frac{21}{24}$ .

Again considering the fourth set of fraction:

LCM of 10 and 15 is 30.Now, multiply by 3 to both the numerator and denominator to $\frac{3}{10}$ . and multiply by 2 to both the numerator and denominator to $\frac{4}{15}$ .We, get the new set of fraction as $\frac{9}{30}\ and \ \frac{8}{30}$ .

∴ $\frac{9}{30}\ >\frac{8}{30}$

6 0

3 years ago

Drove 260 km using 12 liters of gas at this rate how many liters of gas would he need to drive 286km

sdas [7]

Answer:

We calculate the amount of gas used per km and multiply it by 286

Step-by-step explanation:

If we drive for 260km and use 12 litres of gas that means that we use

$\frac{12}{260}$ litres per km.

Now we multiply it by 286:

$\frac{12}{260} *286 = \frac{858}{65} = 13.2$

So, the answer is 13.2 litres.

4 0

3 years ago