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3 years ago

252/9=378/?what's the awnser to this problem

Mathematics

1 answer:

Elina [12.6K]3 years ago

4 0

This equation is false, it has no solutions.

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Seriously question please answer I need it

max2010maxim [7]

Answer:

1. 16cm

2. 26dm

3. 20mm

4. 30dm

5. 78cm

Step-by-step explanation:

1. Perimeter of the square = 4cm + 4cm + 4cm + 4cm = 16cm

2. Perimeter = 8dm + 8dm + 5dm + 5dm

= 26dm

3. Perimeter of rectangle = 3mm + 3mm + 7mm + 7mm = 20mm

4. Perimeter of triangle = 8dm + 12dm + 10dm = 30dm

5. Perimeter = 23cm + 23cm + 16cm + 16cm = 78cm

<h3>Hope it is helpful...</h3>

6 0

3 years ago

How to solve all fraction algebra?

kap26 [50]

Answer:

$- \frac{1}{5} - \frac{4}{9} a = \frac{2}{15}$

$- \frac{4}{9} a = \frac{2}{15} + \frac{1}{5} \\ \\ - \frac{4}{9} a = \frac{1}{3 } \\ \\ ( - 4 \times 3)a = (1 \times 9) \\ - 12a = 9 \\ a = - \frac{9}{12} \\ \\ { \underline{ \: \: a = - \frac{3}{4} \: \: }}$

7 0

3 years ago

What is the circumference and area of a pool with a 30 feet diameter

-Dominant- [34]

Diameter = 30
radius = 15
area = pi x 15 x 15 = 225pi
circumference = 30pi

8 0

4 years ago

Find the area of a triangle with a base of 8 cm and a height of 10 cm

Alexandra [31]

Answer:

Step-by-step explanation:

The formula of an area of a triangle:

$A_\triangle=\dfrac{b\cdot h}{2}$

<em>h</em><em> - height</em>

<em />

We have <em>b = 8cm, h = 10cm</em>.

Substitute:

$A=\dfrac{(8)(10)}{2}=\dfrac{80}{2}=40\ (cm^2)$

6 0

3 years ago

Find the derivative r '(t) of the vector function r(t). t sin 6t , t2, t cos 7t Part 1 of 4 The derivative of a vector function

stich3 [128]

Answer:

$r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))$

Step-by-step explanation:

We need to find the derivative $r'(t)$ of the vector function :

$r(t)=(tsin(6t),t^{2},tcos(7t))$

In order to find $r'(t)$ , we are going to differentiate each of its components ⇒

We can write the following ⇒

$r(t)=(f(t),g(t),h(t))=(tsin(6t),t^{2},tcos(7t))$ ⇒

$f(t)=tsin(6t)\\g(t)=t^{2}\\h(t)=tcos(7t)$

Let's differentiate each function to obtain $r'(t)$ :

$f(t)=tsin(6t)$ ⇒ $f'(t)=1.sin(6t)+t.cos(6t).6=sin(6t)+6tcos(6t)$ ⇒

$f'(t)=sin(6t)+6tcos(6t)$

Now with $g(t)$ :

$g(t)=t^{2}$ ⇒

$g'(t)=2t$

With $h(t)$ :

$h(t)=tcos(7t)$ ⇒ $h'(t)=1.cos(7t)+t[-sin(7t)].7$ ⇒

$h'(t)=cos(7t)-7tsin(7t)$

Finally we need to complete $r'(t)=(f'(t),g'(t),h'(t))$ with its components :

$r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))$

7 0

3 years ago