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2 years ago

Find each number 60% of 45

Mathematics

1 answer:

Sholpan [36]2 years ago

5 0

Answer:

Step-by-step explanation:

$60\%\ of\ 45\\=\frac{60}{100} *45\\=\frac{60}{20}*9\\=3*9\\=27$

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X-5/50°

dusya [7]

Answer:

<QRT = 130°

X = 135

<QRS = 180°

Step-by-step explanation:

1. A straight line is 180°.

180° - 50° = 130°

<QRT = 130°

2. x - 5 = 130°

130° + 5° = 135°

3. As stated before, a straight line is 180°.

3 0

2 years ago

One neckace cost $32.00 and another costs $28.00 sales tax is 8 %.find total cost including tax

inessss [21]

Hi there! So first off, let's add the prices of the two necklaces. 32 + 28 is 60. Now, let's plug in the tax. The tax rate is 8%. Let's add that to 100%, because 100% represents the whole cost. 100 + 8 is 108. That's 108%. We will multiply that percentage by the cost. 60 * 108% (1.08) is 64.8. There. The total cost is $64.80.

3 0

2 years ago

The actual distance between A and B is 960 metres.

Afina-wow [57]

Answer:

Step-by-step explanation:

a. 960 / 4 = 240 for 1 grid

1/2 cm = 24000

1 : 48000

b. A to C

AC² = 3² + 6²

AC² = 9 + 36

AC²= 432 + 1728

AC² = 2160

AC = √2160 km

7 0

2 years ago

What can 45/8 be simplified down to?

Alenkinab [10]

Eight goes into 45, 5 times so it would be 5 and 5/8.

5 0

2 years ago

The triangles below are similar. Triangle A B C. Side A C is 10 and side A B is 5. Angle C is 30 degrees. Triangle D E F. Side E

Gnoma [55]

Answer:

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$\triangle CBA {\displaystyle \sim } \triangle FED$

$\triangle BAC {\displaystyle \sim } \triangle EDF$

Step-by-step explanation:

Given

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$AC = 10$

$AB = 5$

$\angle C = 30^\circ$

$ED = 7.5$

$DF = 25$

$\angle F =30$

$\angle E =90$

Required

Which of the options is/are true:

$\triangle CBA {\displaystyle \sim } \triangle FED$ $\triangle CBA {\displaystyle \sim } \triangle FDE$ $\triangle BAC {\displaystyle \sim } \triangle EFD$

$\triangle BAC {\displaystyle \sim } \triangle EDF$ $\triangle ABC {\displaystyle \sim } \triangle DE\ F$ $\triangle ABC {\displaystyle \sim } \triangle DFE$

The given triangles, implies that:

$A {\displaystyle \sim } \ D$

$B {\displaystyle \sim } \ E$

$C {\displaystyle \sim } \ F$

By taking each sides of both triangle, one after the other; the possible similar triangles are:

$\triangle ABC {\displaystyle \sim } \triangle DE\ F$

$\triangle CBA {\displaystyle \sim } \triangle FED$

$\triangle BAC {\displaystyle \sim } \triangle EDF$

5 0

2 years ago