PLEASE HELP ME , WILL MARK BRAINLIEST

A bag of chocolates is labeled to contain 0.384 pound of chocolates. The actual weight of the chocolates is 0.3798 pound. Are th

taurus [48]

Answer:

The answer to your question is the chocolates are slightly lighter than the weight stated on the label.

Step-by-step explanation:

Data

The label says 0.384 pounds

Actual mass = 0.3798 pounds

Process

To solve this problem the only thing we must do is substrate the actual weight of chocolates to the weight of chocolates stated on the label.

If the result is positive, then the chocolates are lighter

If the result is negative, then the chocolates are heavier

0.384 pounds - 0.3798 pounds = 0.0042

hmu if you need sumin els!

5 0

3 years ago

Read 2 more answers

4y-x=5+2y; 3x+7y=24 solve systen by using substitution

jek_recluse [69]

4y - x = 5 + 2y ..... (1)
3x + 7y = 24 ..... (2)

by grouping like terms in (1)
4y - x = 5 + 2y
4y - 2y - x = 5
<span>-x + 2y = 5 </span> ..... (1a)

by multiplying (1a) through by -3
(-3)(-x) + 2(-3)y = 5(-3)
3x - 6y = -15 ..... (1b)

by subtracting 1a from 2
3x -3x + 7y - (-6y) = 24 - (-15)
13y = 39
⇒ y = 3

by substituting y=3 into (2)
3x + 7(3) = 24
3x = 24 - 21
3x = 3
⇒ x = 1

∴ solution to the system is x=1 when y = 3

6 0

3 years ago

Can you please help me

IRISSAK [1]

Answer:

Step-by-step explanation:

1. HI // KL , J is midpoint of HL 1. Given

2. ∠IHJ ≅ ∠JLK 2. Alternate interior angles are congruent

3. ∠IJH ≅ ∠KJL 3. Vertically opposite angles

4. HJ ≅ JL 4. Given

5. ΔHIJ ≅ ΔLKJ 5. A S A congruent

(Angle Side Angle)

3 0

2 years ago

The areas of the two watch faces have a ratio of 16:25 What is the ratio of the radius of the smaller watch face to the radius o

Leona [35]

Answer:

The ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.

Step-by-step explanation:

Let the Area of smaller watch face be $A_1$

Also Let the Area of Larger watch face be $A_2$

Also Let the radius of smaller watch face be $r_1$

Also Let the radius of Larger watch face be $r_2$

Now given:

$\frac{A_1}{A_2} =\frac{16}{25}$

We need to find the ratio of the radius of the smaller watch face to the radius of the larger watch face.

Solution:

Since the watch face is in circular form.

Then we can say that;

Area of the circle is equal 'π' times square of the radius 'r'.

framing in equation form we get;

$A_1 = \pi {r_1}^2$

$A_2 = \pi {r_2}^2$

So we get;

$\frac{A_1}{A_2}= \frac{\pi {r_1}^2}{\pi {r_2}^2}$

Substituting the value we get;

$\frac{16}{25}= \frac{\pi {r_1}^2}{\pi {r_2}^2}$

Now 'π' from numerator and denominator gets cancelled.

$\frac{16}{25}= \frac{{r_1}^2}{{r_2}^2}$

Now Taking square roots on both side we get;

$\sqrt{\frac{16}{25}}= \sqrt{\frac{{r_1}^2}{{r_2}^2}}\\\\\frac{4}{5}= \frac{r_1}{r_2}$

Hence the ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.

7 0

3 years ago

The diagram above shows a rectangle inscribed in a circle AB=10 and AC =12 caculate the total surface area of the shaded part

erastova [34]

Answer:

$71.63 \: \: \mathrm{cm^2 }$

Step-by-step explanation:

Once we know the diameter of the circle, we can figure out the problem.

The diameter of the circle = The diagonal of the rectangle inscribed in the circle

To find the diagonal of the rectangle, we can use a formula.

$d=\sqrt{w^2 + l^2}$

The width is 10 cm and the length is 12 cm.

$d=\sqrt{10^2 + 12^2}$

$d \approx 15.62$

The diagonal of the rectangle inscribed in the circle is 15.62 cm.

The diameter of the circle is 15.62 cm.

Find the area of the whole circle.

$A=\pi r^2$

The $r$ is the radius of the circle, to find radius from diameter we can divide the value by 2.

$r = \frac{d}{2}$

$r=\frac{15.62}{2}$

$r=7.81$

Let’s find the area now.

$A=\pi (7.81)^2$

$A \approx 191.625$

Find the area of rectangle.

$A=lw$

Length × Width.

$A = 12 \times 10$

$A=120$

Subtract the area of the whole circle with the area of rectangle to find area of shaded part.

$191.625-120$

$71.625 \approx 71.63$

5 0

3 years ago