All categories

3 years ago

Write a problem that involves changing kilograms to grams explain how to find the solution

Mathematics

2 answers:

Serjik [45]3 years ago

8 0

1 kilogram=1,000 grams

GaryK [48]3 years ago

3 0

Like any problem dealing with dimensional analysis. You set up multiplication/division of relationships so that unit labels that you do not need for the final answer cancel out and introduce unit labels that you do need for the final answer...

1kg(1000g/kg)=1000g Notice how equal unit labels cancel out just like equal numbers cancel out. This process makes even the most complicated conversions very simple.

You might be interested in

What is the solution to this equation? 6(x-4)+7=-5

EleoNora [17]

Answer:

x = 2

Step-by-step explanation:

6(x-4) + 7 = -5

6(x-4) + 7 - 7 = -5 - 7

6(x-4) = -12

6(x-4)/6 = -12/6

x - 4 = -2

x - 4 + 4 = -2 + 4

x = 2

4 0

4 years ago

In the right ∆ABC, the hypotenuse AB = 17 cm. M is the midpoint of the hypotenuse. Find the legs if PAMC=32 cm and PBMC=25 cm

jeyben [28]

Answer:

The length of the legs is 8.64cm and 14.64cm respectively

Step-by-step explanation:

I've added an attachment to aid my explanation.

At different intervals, I'll be making reference to it.

Given

$AB = 17$

$PAMC = 32$

$PBMC = 25$

From the attachment, we have:

$y + z = AB$

Since, M is the Midpoint

$y = z = \½AB$

Substitute 17 for AB

$y = z = \½ * 17$

$y = z = 8.5$

Also, from the attachment

$v + x + z = PAMC$

$v + x + y = 32$

Substitute 8.5 for y

$v + x + 8.5 = 32$

$v + x = 32 - 8.5$

$v + x = 23.5$ --------- (1)

Also, from the attachment

$v + w + z = 25$

Substitute 8.5 for z

$v + w + 8.5 = 25$

$v + w = 25 - 8.5$

$v + w = 17.5$ ----------- (2)

Subtract (2) from (1)

$v - v + x - w = 23.5 - 17.5$

$x - w = 6$

Make x the subject

$x = 6 + w$

Apply Pythagoras Theorem:

We have that:

$AB^2 = AC^2 + BC^2$

The above can be replaced with

$17^2 = x^2 + w^2$ (see attachment)

$289 = x^2 + w^2$

Substitute 6 + w for x

$289 = (6 + w)^2 + w^2$

$289 = 36 + 12w + w^2 + w^2$

$289 - 36 = 12w + 2w^2$

$253 = 12w + 2w^2$

Reorder

$2w^2 + 12w - 253 = 0$

Solve using quadratic equation:

$w = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where

$a = 2$

$b = 12$

$c = -253$

$w = \frac{-12 \± \sqrt{12^2 - 4 * 2 * -253}}{2 * 2}$

$w = \frac{-12 \± \sqrt{144 + 2024}}{4}$

$w = \frac{-12 \± \sqrt{2168}}{4}$

$w = \frac{-12 \± 46.56}{4}$

Split:

$w = \frac{-12 + 46.56}{4}$ or $w = \frac{-12 - 46.56}{4}$

$w = \frac{34.56}{4}$ or $w = \frac{-58.56}{4}$

$w = 8.64$ or $w = -14.64$

But length can't be negative

So:

$w = 8.64$

Recall that: $x = 6 + w$

$x = 6 + 8.64$

$x = 14.64$

<em>Hence, the length of the legs is 8.64cm and 14.64cm respectively</em>

5 0

3 years ago

Need help with this ignore what a put in the box it isn’t right

Zarrin [17]

$f=6cm\\g=8cm$

Why?

The first thing we need to do is find the area of the triangle, we can to that by subtracting the area of ABCD from ACBE, then, we can use the formulas to calculate the area for both triangle and rectangle to find "f" and "g".

Calculating we have:

$TriangleArea=ABCE-ABCD\\\\TriangleArea=60cm^{2}-48cm^{2}=12cm^{2}$

Now, we can calculate "f" by using the formula to calculate the area of the triangle:

$TriangleArea=\frac{b*h}{2}\\\\TriangleArea=\frac{f*4cm}{2}\\\\12cm^{2}*2=f*4cm\\\\\frac{24cm^{2}}{4cm}=f\\\\f=6cm$

Now, finding "g" by using the formula to calculate the area of the rectangle, we have:

$RectangleArea=ABCD\\\\ABCD=Base*Height\\\\48cm^{2}=base*6cm\\\\base=g=\frac{48cm^{2}}{6cm}=8cm$

Hence, we have that:

$f=6cm\\g=8cm$

Have a nice day!

8 0

3 years ago

What is -(x + 5) distributed

KiRa [710]

Answer:

-x-5

Step-by-step explanation:

Multiply every part inside the equation by -1.

x*-1 and 5*-1

-1x+-5x

Simplify.

-x-5

3 0

3 years ago

Find an equation of the plane. the plane through the point (−4, 8, 10) and perpendicular to the line x = 5 + t, y = 2t, z = 1 −

Allushta [10]

<span>1) Find an equation of the plane. The plane that passes through the point (2, 3, 4) and contains the line x = 4t, y = 2 + t, z = 3 − t 2) Find an equation of the plane. The plane that passes through (6, 0, −3) and contains the line x = 2 − 4t, y = 1 + 5t, z = 2 + 2t 3) Find an equation of the plane. The plane that passes through the point (1, −1, 1) and contains the line with symmetric equations x = 2y = 5z 4) Find the point at which the line intersects the given plane. x = 2 − t, y = 1 + t, z = 3t; x − y + 5z = 14 5) Find the point at which the line intersects the given plane. x = 2 + 2t, y = 3t, z = 4 − 2t; x + 2y − z + 2 = 0 6) Find the point at which the line intersects the given plane. x = y − 2 = 2z; 2x − y + 2z = 2</span>

6 0

3 years ago