All categories

3 years ago

The mass of a chocolate bar is 200 g. A box contains 150 chocolate bars. What is the

Mathematics

2 answers:

Veseljchak [2.6K]3 years ago

4 0

Answer:

30kg

Step-by-step explanation:

mass of 1 Chocolate bar = 200g

Box = 150 chocolate bar

150x200g = 30,000g

1kg =1000g

so 30,000g/1000g x 1kg

= 30kg

Luba_88 [7]3 years ago

4 0

I got 30,000 when I multiplied 200 time 15”. Then I searched in google for 30,000 grams transferee into kilo grams and it’s 30! Hope this helps! Give me a like please it would be appreciated

You might be interested in

12/39 in simplest form is 4/13 what are two ways to get there

arlik [135]

Divide 3 on the numerator and denominator.

Hoped this helped.

~Bob Ross®

7 0

3 years ago

Read 2 more answers

Subtract 43 from 20 raised to the 7th power; then multiply by 3.

viktelen [127]

Hi there! :)

Answer:

Subtract 43 from 20 raised to the 7th power; then multiply by 3. The answer is: 3,839,999,871

Step-by-step explanation:

"Subtract" means to take away a number from another.

" 20 raised to the 7th power" is the same thing as 20 exponent 7: $20^{7}$

SO, you want to take away 43 from $20^{7}$ and then multiply the answer by 3:

( $20^{7}$ - 43) × 3

$20^{7} =$ 1,280,000,000

(1,280,000,000 - 43) × 3

1,280,000,000 - 43 = 1,279,999,957

1,279,999,957 × 3

1,279,999,957 × 3 = 3,839,999,871

There you go! I really hope this helped, if there's anything just let me know! :)

6 0

3 years ago

Read 2 more answers

Two cables support a 800-lb weight, as shown. Find the tension in each cable.

jeka94

Answer:

892 lb (right)
653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

Rcos(45°) +Lcos(75°) = 800

Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

= 800sin(75°)/sin(120°) ≈ 892 . . . pounds

L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

_____

This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

(right, left) = w/sin(α+β)×(sin(β), sin(α))

5 0

3 years ago

Which transformations will produce similar, but not congruent, figures?

atroni [7]

Answer:

The correct answers are A B D

Step-by-step explanation:

hope this helped!

4 0

2 years ago

The coordinates of AB are A(-5, -1) and B(-2, -6). If AC : CB = 3 : 7, what are the coordinates of point C?

navik [9.2K]

$\text{Let the coordinates of AB are }A(-5,-1), \text{ and }B(-2, -6)\\
\\
\text{let the coordinate of th point C be (x, y), such that}\\
\\
AC:CB=3:7\\
\\
\text{By the section formula, we know that if P(x,y) lies on a line segment AB}\\
\text{where, }A(x_1,y_1), \ B(x_2,y_2)\text{ and satisfy the ratio, }AP:PB=m:n, \text{ then}$

$P=\left ( \frac{mx_2+nx_1}{m+n}, \ \frac{my_2+ny_1}{m+n} \right ).\\
\\
\text{So using this formula, we get for the given problem}\\
\\
C=\left ( \frac{3(-2)+7(-5)}{3+7}, \ \frac{3(-6)+7(-1)}{3+7} \right )\\
\\
\Rightarrow C=\left ( \frac{-6-35}{10}, \ \frac{-18-7}{10} \right )\\
\\
\Rightarrow C=\left ( \frac{-41}{10}, \ \frac{-25}{10} \right )\\
\\
\Rightarrow C=\left ( -4.1, \ -2.5 \right )$

Coordinates of the point C is: (-4.1, -2.5)

5 0

3 years ago