Ask question

Ask question

All categories

3 years ago

11

A right triangle has a vertex at point M and a height of 3 units. The base of the triangle is on MN and is 4 units long. Use the

polygon tool to draw the triangle. Each segment on the grid represents 1 unit.

1 answer:

Ipatiy [6.2K]3 years ago

7 0

Answer:

I used the answer the other person gave, but it said it was wrong. Here was the answer I put in the second time.

Step-by-step explanation:

You might be interested in

PLS HELP ASAP (SHOW WORK) LOTS OF POINTS + give thanks if CORRECT

GuDViN [60]

Multiply by two
2/5 x 2/1 = 4/5

3 0

3 years ago

B(14, 12)<br> A(-5,-2)<br> A. Find the midpoint of Segment AB

IRISSAK [1]

The midpoint formula is (x1+x2/2, y1+y2/2
so you would do (-5+14)/2=4.5. (-2+12)/2=5
(4.5, 5)

3 0

3 years ago

Drag each tile to the correct box.

Anni [7]

Answer:

$3x^{2}y^{2}\sqrt[4]{7xy^{3}}$ .

Step-by-step explanation:

Since we have to find the simplified form of the expression as below

$\sqrt[4]{567x^{9}y^{11}}$

we will solve the expression as below

$=[567x^{9}y^{11}]^{\frac{1}{4}}$

$=(81\times 7)^{\frac{1}{4}}\times x^{\frac{9}{4}}\times y^{\frac{11}{4}}$

$=81^{\frac{1}{4}}.7^{\frac{1}{4}}.x^{\frac{8}{4}+\frac{1}{4}}.y^{\frac{8}{4}+\frac{3}{4}}$

$=(3^{4})^{\frac{1}{4}}.7^{\frac{1}{4}}.x^{2+\frac{1}{4}}.y^{2+\frac{3}{4}}$

$=3^{1}.7^{\frac{1}{4}}.x^{2}.x^{\frac{1}{4}}.y^{2}.y^{\frac{3}{4}}$

$=3.x^{2}.y^{2}.(7^{\frac{1}{4}}.x^{\frac{1}{4}}.y^{\frac{3}{4}})$

$=3x^{2}y^{2}.(7xy^{3})^{\frac{1}{4}}$

$=3x^{2}y^{2}\sqrt[4]{7xy^{3}}$ .

7 0

3 years ago

40% lower than 1,110

maks197457 [2]

Answer:

its 444

Step-by-step explanation:

6 0

3 years ago

Mandy is thinking of a prime number that is greater than 41 and less than 47 .What is Mandy's number

kipiarov [429]

Mandy's number is 43

7 0

3 years ago

Read 2 more answers