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3 years ago

Which shows all the names that apply to the figure

Mathematics

1 answer:

ololo11 [35]3 years ago

5 0

Parallelogram rhombus and sometimes rectangle

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Kevin drew a diagonal inside a quadrilateral. He realizes the it always make 2 congruent isosceles triangleso. What kind of quad

olya-2409 [2.1K]

Answer:

The square is the quadrilateral.

Step-by-step explanation:

A quadrilateral is a 2 dimensional geometric shape. It has four sides which may be or may not be equal.

If a quadrilateral is a square and we make its diagonal so that we get two triangles.

These two triangles are isosceles triangle as the two sides are equal and the square has four equal sides and the triangles are congruent to each other.

here, the triangles ABC and ADC are congruent and isosceles.

7 0

3 years ago

Find the surface area.<br> 7 in.<br> 7 in.<br> 7 in.<br> please help

saveliy_v [14]

Answer:

294

Step-by-step explanation:

One side is 7×7 which is 49.

There are six sides so now it's 49×6, which gives you 294.

6 0

2 years ago

What's the area of this? PLS HELP

balandron [24]

Answer:

446mm

Step-by-step explanation:

If we box off parts of the area, it makes it easier to solve. I personally broke it into tiny bits:

Upper left box: 16mm

Bigger box (excluding little box): 90mm

Big rectangle: 340mm

Now, add them all together.

Equals 446mm

3 0

2 years ago

Read 2 more answers

Find the midpoint of the line segment joining the points R(-5,5) and S(4.6).

tigry1 [53]

Hey there! I'm happy to help!

To find the x value of the midpoint, you add the x values and divide by 2.

-5+4=-1

-1/2=-1/2

For the y value, you add the y values and divide by 2.

5+6=11

11/2=5 1/2

So, the midpoint is (-1/2, 5 1/2).

Have a wonderful day! :D

6 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago