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3 years ago

5

Kenya went to the convenience store to buy some candy bars for her friends. She spent a total of $6.00 on candy bars. Each candy

bar cost $1.50.
How many candy bars did Kenya buy?

2 answers:

iris [78.8K]3 years ago

4 0

Answer:

4

Step-by-step explanation:

lawyer [7]3 years ago

4 0

Answer:

4

Step-by-step explanation:

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PLEASE HELP ASAP SEE THE PICTURE!!!!

nata0808 [166]

The number that Samya can be thinking of is 571.

<h3>How to get the number?</h3>

The information stated that Samya was thinking of a number that's is between 560 and 590. It was further stated that the number isn't a multiple of four and that the addition of it's digit will give a prime number.

This will be:

5 + 7 + 1

= 13

In this case, 13 is a prime number. Also, 571 is not a perfect square.

Therefore, the number is 571.

Learn more about multiples on:

brainly.com/question/26856218

#SPJ1

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9 months ago

If he starts at 1 and jumps 3 units to the right, then where is he on the number line? How far away from zero is he? (explain ho

GarryVolchara [31]

Answer:

4, he is 4 spaces away from zero.

Step-by-step explanation: I know things

7 0

3 years ago

Find the vertex and length of the latus rectum for the parabola. y=1/6(x-8)^2+6

Ivan

Step-by-step explanation:

If the parabola has the form

$y = a(x - h)^2 + k$ (vertex form)

then its vertex is located at the point (h, k). Therefore, the vertex of the parabola

$y = \dfrac{1}{6}(x - 8)^2 + 6$

is located at the point (8, 6).

To find the length of the parabola's latus rectum, we need to find its focal length <em>f</em>. Luckily, since our equation is in vertex form, we can easily find from the focus (or focal point) coordinate, which is

$\text{focus} = (h, k +\frac{1}{4a})$

where $\frac{1}{4a}$ is called the focal length or distance of the focus from the vertex. So from our equation, we can see that the focal length <em>f</em> is

$f = \dfrac{1}{4(\frac{1}{6})} = \dfrac{3}{2}$

By definition, the length of the latus rectum is four times the focal length so therefore, its value is

$\text{latus rectum} = 4\left(\dfrac{3}{2}\right) = 6$

5 0

3 years ago

Vanessa draws one side of equilateral ΔABC on the coordinate plane at points A(–2, 1) and B(4,1). What are the possible coordina

grandymaker [24]

Important question. No menu of choices given so we'll just have to figure it out.

It's not possible in two dimensions for all three vertices of an equilateral triangle to be lattice points, i.e. points with integer coordinates.

Since A(-2,1) and B(4,1) have integer coordinates, our other point won't. There will be a √3 involved, the algebraic tell that there's a 30/60/90 triangle lurking somewhere in the problem.

Here the given side is parallel to the x axis which makes things easier.

|AB| = 4 - -2 = 6

The x coordinate of the third vertex will be the same as that of the midpoint of AB, let's call it M,

M = ( (-2 + 4) / 2, (1 + 1)/2 ) = (1, 1)

We're making some progress. Whatever our height h is our two candidates for the third vertex C are (1, 1 ± h).

Since |AB|=6, we get |AB|=|BC|=|AC|=6 because it's an equilateral triangle.

The altitude CM bisects the triangle into 2 30/60/90 triangles. Let's take one of them, AMC. Angle AMC is a right angle, so we have a right triangle with legs |AM|=3 and |CM|=h, and hypotenuse |AC|=6. The Pythagorean Theorem tells us:

3² + h² = 6²

9 + h² = 36

h² = 27

h = 3√3

Answer: C is (1, 1 + 3√3) or (1, 1 - 3√3)

4 0

3 years ago

The 6 in the number 6,140 has a value __-- 648. times greater than the 6 in the number

nevsk [136]

Answer:

1,000 because it's in the thousands place

Step-by-step explanation:

6 0

3 years ago