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3 years ago

About how many pounds are in 14 kilograms?

Mathematics

2 answers:

sesenic [268]3 years ago

6 0

Answer:

308.647

Step-by-step explanation:

Its searchable pls mark brainliest.

lys-0071 [83]3 years ago

4 0

Answer:

14 kg = 30.965 pounds

Step-by-step explanation:

for an approximate result, multiply the mass value by 2.205

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Please help I really need help on this I would really appreciate it anything helps !

andreev551 [17]

Answer:

AXE and CXD are vertical angles

AXF and FXD are supplementary angles

DXC and BXC are complementary angles

EXA and AXB are adjacent angles

AXC and CXD are supplementary angles

EXD and AXC are vertical angles

Step-by-step explanation:

6 0

3 years ago

In 2010, the population of a city was 161,000. From 2010 to 2015, the population grew by 8%. From 2015 to 2020, it fell by 6.7%.

solniwko [45]

Answer:

99%

Step-by-step explanation:

2010 population = 161,000

From 2010 to 2015, the population grew by 8%.

2010 to 2015 = 161,000 + (8% of 161,000)

= 161,000 + (0.08 * 161,000)

= 161,000 + 12,880

= 173,880

2010 to 2015 = 173,880

From 2015 to 2020, it fell by 6.7%

2015 to 2020 = 173,880 - (6.7% of 173,880)

= 173,880 - (0.067 * 173,880)

= 173,880 - 11,649.96

= 162,230.04

2015 to 2020 = 162,230.04

what percent did the city grow from 2010 to 2020.

= 161,000 / 162,230.04 × 100

= 0.9924179270374 × 100

= 99.241792703743

Approximately, 99% to the nearest whole number

4 0

2 years ago

Describe how the graph of y= x2 can be transformed to the graph of the given equation. y = x2 + 3 (5 points)

kari74 [83]

Answer:

x=-1 1/2,x=-1.5

Step-by-step explanation: x=-3/2

6 0

3 years ago

What value of a makes the equation true? 1.7a+0.3a=4/5 btw 4/5 is a fraction

andreyandreev [35.5K]

1.7a + 0.3a = 0.8;
(1.7 + 0.3)a = 0.8;
2 x a = 0.8;
a = 0.8 ÷2;
a = 0.4;

5 0

3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago