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nikitadnepr [17]
2 years ago
13

. XYZ has vertices X (–5, 2), Y (0, –4), and Z (3, 3). What are the vertices of the image of X’Y’Z’ under the translation (x, y)

→ (x + 7, y – 5) ?
Mathematics
1 answer:
Katen [24]2 years ago
6 0

Answer:

X' = (2, -3)

Y' = (7, -9)

Z' = (10, -2)

Step-by-step explanation:

to find the translation of each point substitute the x and y with the the numbers

eg; X ( -5,2) the x is 5 and the y is 2

so -5 + 7 = 2 (the x value)

2 - 5 = -3 ( the y value)

You might be interested in
3. i) Determine the volume of the cylinder above to the nearest cubic centimetre. (3 points)​
klemol [59]

Answer:

1480cm³

Step-by-step explanation:

The Equation for the Volume of a Cylinder is V=Bh, where B is the area of the circle (A= πr²).

We are given all of the things we need, we just have to plug them into the equation.

First let's find B using the radius(r) which is given to be 6.4cm, so B=πr²

B=π(6.4cm)²

Plug that into a calculator to get 128.6796351cm²

Now we can plug that into the original equation, V=Bh.

V=(128.6796351cm²)h

h is also given to be 11.5cm, so lets plug that in.

V=(128.6796351cm²)(11.5cm)

You will get V=1479.815804cm³, but since we are rounding to the nearest cubic centimeter, we would round up to 1480cm³.

3 0
3 years ago
Can you answer these please many thanks
givi [52]

Given: (a) 2(x+3)=2x+6 and (b) 4(y-3)=4y-12.

To find: The expanded form of the given expressions.

(c) 4(m+n)=4m+4n       (using a(b+c)=ab+ac)

(d) 3(5-q)=15-3q           (using a(b-c)=ab-ac)

(e) 5(2c+1)=10c+5          (using a(b+c)=ab+ac)

(f) 3(2x-5)=6x-15          (using a(b-c)=ab-ac)

(g) 7(4b-1)=28b-7          (using a(b-c)=ab-ac)

(h) 3(2x+y-5)=3((2x+y)-5)

⇒3(2x+y-5)=3(2x+y)-15         (using a(b-c)=ab-ac)

⇒3(2x+y-5)=6x+3y-15            (using a(b+c)=ab+ac)

(i) 2(6a-4b+3)=2((6a-4b)+3)

⇒2(6a-4b+3)=2(6a-4b)+6         (using a(b+c)=ab+ac)

⇒2(6a-4b+3)=12a-8b+6            (using a(b-c)=ab-ac)

(j)6(m+n+p)=6((m+n)+p)

⇒ 6(m+n+p)=6(m+n)+6p           (using a(b+c)=ab+ac)

⇒ 6(m+n+p)=6m+6n+6p            (using a(b+c)=ab+ac)

(k) y(y+2)=y^2+2y          (using a(b+c)=ab+ac)

(l) g(g-3)=g^2-3g           (using a(b-c)=ab-ac)

(m) n(4-n)=4n-n^2        (using a(b-c)=ab-ac)

(n) a(b+c)=ab+ac           (using a(b+c)=ab+ac)

(o) s(3s-4)=3s^2-4s        (using a(b-c)=ab-ac)

(p) 2x(x+5)=2x^2+10x     (using a(b+c)=ab+ac)

(q) 4y(x-3)=4xy-12y      (using a(b-c)=ab-ac)

(r) 5a(2b-5)=10ab-25a     (using a(b-c)=ab-ac)

(s) 4a(3b+2c)=12ab+8ac    (using a(b+c)=ab+ac)

(t) 5p(4p-5q)=20p^2-25pq    (using a(b-c)=ab-ac)

6 0
3 years ago
Point T is located at (6, –2) and point S is located at (1, 13). What are the coordinates of the point that is 4/5 of the way fr
Darya [45]

Answer:

(2, 10)

Step-by-step explanation:

Making the points on a graph and creating a line between them, you can see the equation to connect the two points is -3x + 16. Then, because the points are 5 units apart horizontally, go to the point that is 4 units left of point T (1/4 of the way between T and S) and that is your point: (2, 10)

I apologize if this is incorrect :)

7 0
3 years ago
Triangle DEF has vertices located at D (2, 1), E (3,5), and F (6,2).
Nataly_w [17]

Answer:

3cm 4cm and 5 cm

Step-by-step explanation:

we can find the accurate answer by plotting in a graph.thank you

3 0
3 years ago
Please answer and explanation
mrs_skeptik [129]

61.23 square inches of metal is needed to create a cylindrical can.

Solution:

Diameter of the base = 3 in

Radius of the base = 3 ÷ 2 = 1.5 in

Height of the cylinder = 5 in

The value of π = 3.14

<u>To find the surface area of the cylinder:</u>

Surface area of the cylinder = 2 \pi r^{2}+2 \pi r h

                                               = 2 × 3.14 × (1.5)² + 2 × 3.14 × 1.5 × 5

                                               = 14.13 + 47.1

Surface area of the cylinder = 61.23 sq. in

Hence 61.23 square inches of metal is needed to create a cylindrical can.

3 0
3 years ago
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