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3 years ago

Given: vu=st, sv=tu prove vx=xt

1 answer:

6 0

Given <span>vu=st, sv=tu

</span>STUV is a parallelogram : If both pairs of opp. sides of a quad. are , then the quad is a parallelogram.

VX, XT = The diagonals of a parallelogram bisect each other.

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Answer the geometry question in the pic and save my life...

Anastasy [175]

Answer:

The answer should be A'(-9,3) B'(12,-6)

Step-by-step explanation:

If you dialate, you multiply every coordinate by 3. -3 x 3= -9. 1 x 3=3. 4 x 3= 12. -2 x 3= -6. So, the last option is correct.

3 0

3 years ago

Find the value of the missing coefficient in the factored form of 27f^3 + 125g^3. 27f^3+125g^3=(3f+5g)(9f^2-?fg + 25g)^2

7nadin3 [17]

Answer:

Step-by-step explanation:

The formula for factoring a sum of cubes is:

$a^3+b^3=(a+b)(a^2-ab+b^2)$

We have a=3f and b=5g here.

So a*b in this case is 3f*5g=15fg.

The ? is 15.

6 0

2 years ago

Determine the combined surface area of a cube with an edge that is 3.5 cm long and a cube with an edge that is 2 cm long.

Elis [28]

Answer:

97.5

Step-by-step explanation:

3.5cm cube surface area:

3.5*3.5*6 (the 6 faces of the cube) =

73.5

2cm cube surface area:

2*2*6 =

combined surface area:

73.5+24=

97.5

5 0

2 years ago

Answer #5,6,7,8,9,10

MatroZZZ [7]

$5)\ -2\times\frac{3}{7}=-\frac{2\times3}{7}=-\frac{6}{7}\\\\6)\ -2\frac{2}{3}\times4\frac{1}{10}=-\frac{8}{3}\times\frac{41}{10}=-\frac{4}{3}\times\frac{41}{5}=-\frac{164}{15}=-10\frac{14}{15}\\\\7)\ -2\frac{1}{5}\times(-1\frac{3}{4})=-\frac{11}{5}\times(-\frac{7}{4})=\frac{77}{20}=3\frac{17}{20}\\\\8)\ -1\frac{1}{4}\times9=-\frac{5}{4}\times9=-\frac{45}{4}=-11\frac{1}{4}\\\\9)\ -1\frac{5}{7}\times(-2\frac{1}{2})=-\frac{12}{7}\times(-\frac{5}{2})=\frac{6}{7}\times\frac{5}{1}=\frac{30}{7}=4\frac{2}{7}$

$1)\ -2\frac{3}{8}\times2\frac{1}{2}=-\frac{19}{8}\times\frac{5}{2}=-\frac{95}{16}=-5\frac{15}{16}$

7 0

2 years ago

Read 2 more answers

Can someone help me pls

marysya [2.9K]

Answer its a

Step-by-step explanation:

7 0

3 years ago