Find three consecutive integers whose sum is six times the greater integer

Which verbal expression best describes the algebraic expression 2x ÷ 4?

Vlada [557]

Let's reason through the answers...

A) Do you see any sums? Any addition? Me neither... it's not A.

B) Well I see a product of 2 and "some number," but are we dividing by 2? Nope. Not B.

C) I see a quotient (2x) divided by 4, but the "some number" is not being multiplied by four. Not C.

D) Well this is the only one left, so it better be this one! We have a quotient, and we have two times some number (2x). The quotient (the division) is of 2x and four. This is it! It's D.

5 0

3 years ago

Theresa’s parents, Cal and Julia, are getting phones that each have 64 gigabytes of storage. How many bits of storage come with

disa [49]

64 divided by 2 =32 gigabytes

I hope this helps you.

7 0

3 years ago

Read 2 more answers

HELP WITH BOTH QUESTIONS SUPER CONFUSED!

Flauer [41]

There's only one question; you are confused.

$\angle CXY \cong \angle BXY \qquad \textrm{Given}$

$\angle CXA \cong \angle BXA \qquad \textrm{Supplements of congruent angles are congruent}$

$\overline{AX} \cong \overline{AX} \qquad \textrm{Reflexivity of congruency}$

$\angle CAX \cong \angle BAX \qquad \textrm{Given}$

$\triangle CAX \cong \triangle BAX \qquad \textrm{ASA}$

$\overline{CX} \cong \overline{BX} \qquad \textrm{Corresponding parts}$

$\overline{XY} \cong \overline{XY} \qquad \textrm{Reflexivity of congruency}$

$\triangle XCY \cong \triangle XBY \qquad \textrm{SAS}$

$\angle XCY \cong \angle XBY \qquad \textrm{Corresponding parts}$

6 0

3 years ago

Please help answer !!

wariber [46]

Answer:

32.3

Step-by-step explanation:

First we want to make a triangle like on the left side, so make a new point on AD that is directly below C, we'll call this E.

We know CE is 6 because it's the same as that line going down from B, let me know if you do not see that.

now we want to find the last leg of this new triangle CED, the non hypotenuse. We know AD is 24 CM. It is now sebarated into three parts, the beginning part in the left most triangle, the 10 cm between the triangles and the side we want to find. Since we know the middle part we want to find the left part.

The left part of AD can be found with the pythagorean theorem. $7.5^2=6^2+x^2$ this can be turned into $\sqrt{7.5^2+6^2}=x=4.5$ Now we know 24 = 10 + 4.5 + y, with y being the right part. So now we know y=9.5

Back to the new triangle, we know two sides and want to find an angle, so we can do some inverse trig functions. Relatie to the angle we want to find we have the opposite (6) and the adjacent (9.5) so we use tangent. tan(theta)= o/a = 6/9.5. Using the inverse of tangent or arctan (because it's easier to type) arctan(6/9.5)=32.276 which rounds to 32.3. Again, let me know if you need help with any of this.

8 0

3 years ago

In ∆ABC with m∠C = 90° the sides satisfy the ratio BC:AC:AB = 4:3:5. If the side with middle length is 12 cm, find: 1) The perim

UNO [17]

Please see the attached picture to see the similar triangle.

The new triangle has triple the side lengths of the first one. This means all side lengths will be triple.

You will use this ratio to find the new side lengths and then use them to find the perimeter and area.

3:4:5 = 9 cm: 12 cm:15 cm on triangleABC

1. The perimeter is 9cm + 12 cm + 15 cm or 36 cm.

2. The area is 1/2 x 12 x 9 or 54 square cm.

3. The length of the hypotenuse is 15 cm.

3 0

3 years ago