What portion of shared $7500 security costs should be apportioned to store A?

An item is regularly priced at $20. Keiko bought it on sale for 15% off the regular price. How much did Keiko pay?

Anna71 [15]

Pay = 20 - (20 x .15) = $17

3 0

4 years ago

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I need to solve and find x. 4x+15<40

antiseptic1488 [7]

Answer:

$x$ or $x< 6 \frac{1}{4}$

Step-by-step explanation:

4x + 15 < 40

Subtract 15 from both sides

4x + 15 - 15 < 40 - 15

4x < 25

Divide both sides by 4

4x/4 < 25/4

x < 6.25

or

x < 6 1/4

3 0

3 years ago

Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral

Stella [2.4K]

Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

$l_{AB} = \sqrt{\left (6-4 \right )^{2}+\left (6-1 \right )^{2}} = \sqrt{29}$

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

$l_{BC} = \sqrt{\left (1-6 \right )^{2}+\left (4-6 \right )^{2}} = \sqrt{29}$

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

$l_{CD} = \sqrt{\left (-1-1 \right )^{2}+\left (-1-4 \right )^{2}} = \sqrt{29}$

The length of side DA, with D = (-1, -1) and A = (1,4) gives;

$l_{DA} = \sqrt{\left (4-(-1) \right )^{2}+\left (1-(-1) \right )^{2}} = \sqrt{29}$

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Therefore;

$Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1$

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

$Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1$

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

5 0

3 years ago

Can someone Please Help

Iteru [2.4K]

Answer:

4

Step-by-step explanation:

6 0

3 years ago

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A 1,200g block of phosphorus-32, which has a half life of 14.3 days, is stored for 100.1 days. At the end of this period, how mu

Valentin [98]