please is this a function and why? My mom already got mad at me I don’t want her getting mad anymore if I get this wrong:(

(07.01 MC)

sweet-ann [11.9K]

Answer:

C

Step-by-step explanation:

5 0

3 years ago

A square has sides 5.3cm long. Work out the area of the square

mezya [45]

Answer:

Area of square = 28.09 square cm.

Step-by-step explanation:

Given : A square has sides 5.3 cm long.

To find : Work out the area of the square.

Solution : We have given

Side of square = 5 .3 cm .

Area of square = side * side .

Area of square = 5.3 * 5.3 .

Area of square = 28.09 square cm.

Therefore, Area of square = 28.09 square cm.

4 0

3 years ago

Helpppppppp asap 89points

Lynna [10]

Answer:

the answer for the first is the second one

and the answer for the second is 9>6

Step-by-step explanation:

9 is greater than (>) six

and when plotted you can see that 9 is to the right of -6

hope i helped!

7 0

2 years ago

Read 2 more answers

Let's say we have a triangle with sides called a, b, and c, and that b is the base. Our triangle has a height that we'll call h.

Irina18 [472]

Answer:

16.34 square inches.

Step-by-step explanation:

Given:

Base of triangle, b = 3 inches

Height of triangle, h = 10.89 inches

Question asked:

What is the area of this triangle in square inches?

Round your answer to the nearest hundredths place .

Solution:

As we know,

$Area\ of\ triangle =\frac{1}{2} \times base\times height$

$=\frac{1}{2} \times3\times10.89\\\\ =\frac{32.67}{2} \\\\ =16.335$

As, we have to find the answer to the nearest hundredths place, hence the area of triangle will be 16.34 square inches.

7 0

3 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago