All categories

2 years ago

Is the enlargement or a reduction? Find the scale factor of the dilation. Explain

Mathematics

1 answer:

malfutka [58]2 years ago

4 0

Good evening ,

Answer:

In this case the dilation is a reduction (as you can see the image triangle is smaller than the original triangle)

and also the scale factor = OA’/OA = 2/6 = 1/3

since -1 < 1/3 < 1 then it’s a reduction.

Step-by-step explanation:

Look at the photo below for more details (notice OA’ and OA).

________________________________________________

You might be interested in

A lake has a small patch of lily pads and every day the patch grows to double its size. It takes 32 days for the patch to cover

S_A_V [24]

Answer:

It took 31 days for the patch to cover half the lake

Step-by-step explanation:

The patch grows to double its size everyday

the patch completely covers the lake in 32 days

Since the patch doubles itself everyday, this means that the previous day before the 32nd day, the lake was just half covered.

Therefore, the the patch covered half the lake on the 31st day, i.e it took 31 days for the patch to cover half the lake

6 0

2 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

2 years ago

Using pemdas please

Bond [772]

$4^{3}+2(8)=80$

Solution:

Given expression:

$4^{3}+2(8)$

Expansion of PEDMAS:

Parenthesis, Exponents, Division, Multiplication, Addition, Subtraction.

To solve the given expression using pedmas rule.

First solve the expression with parenthesis.

$4^{3}+2(8)=4^{3}+16$

Next do the exponents.

$4^{3}+2(8)=64+16$

Finally do the addition.

$4^{3}+2(8)=80$

Hence the answer is 80.

6 0

3 years ago

Read 2 more answers

Work out the area of this semi circle

kirill [66]

Answer:

39.27

Step-by-step explanation:

The area of a semi-circle is pi times the radius squared divided by 2. So basically standard area of a circle formula (pi times the radius squared) but then you add in the extra step of dividing it by 2.

The formula looks like this:

π × r² ÷ 2

The radius is half the diameter of the circle. So basically think of it as a line to the middle of the circle. Our diameter is 10, (the diameter is the entire length of the circle. It gives us this already. If you need diameter and only have the radius add the radius together twice or multiply it by 2 to get the diameter. Fun fact..?)

and it tells us to substitute pi for 3.142.

So, plugging our numbers in we get:

3.142 x 5^2 ÷ 2 = 39.275.

Cut out the 3rd decimal place (5) as it tells us to only include the first 2 decimal places, and we finally end up with 39.27.

Hope this helps and have a nice day!

5 0

2 years ago

A surveyor is conducting measures to see how wide a road is

drek231 [11]

Answer:

D) 45 ft

Step-by-step explanation:

The two triangles are shown below.

Given:

BC = 60 ft, CD = 24 ft and DE = 18 ft.

Since, the two triangles are similar, their corresponding sides are in proportion.

So, $\frac{AB}{DE}=\frac{BC}{CD}=\frac{AC}{CE}$

Now, consider the proportion of sides,

$\frac{AB}{DE}=\frac{BC}{CD}\\AB=\frac{BC}{CD}\times DE\\AB=\frac{60}{24}\times 18\\AB=\frac{60\times 18}{24}=45\ ft$

Therefore, the distance between A and B is 45 ft.

8 0

3 years ago