Does anyone have an answer?

A jewler buys a ring for $85 he sells it for 135% increase what is theretail of the ring

erik [133]

Q: A jeweler buys a ring for $85 he sells it for 135% increase what is the retail of the ring

A: 199.75 just think about the increase of 35% and add 1

6 0

3 years ago

At the moment, a physical therapist has 9 9/12 rolls of medical tape on hand. He anticipates using up 6 6/12 rolls while working

mart [117]

Answer:

subtract 6.5 by 9.75

Step-by-step explanation:

9/12 =.75

6/12=.5

9.75-6.5=3.25

hope this helps

°_°

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

A golf ball is hit with a club from the ground on a flat surface. The height of the ball, h(t), in feet, t seconds after it was

Arlecino [84]

Answer:

The correct numbers are;

1) 0

2) 25

3) 1.25

4) 2.5

When the ball is hit at 0 seconds, it has a height of 0 feet. The ball's height increases until it reaches its maximum height of 25 feet after 1.25 seconds. The ball's height then decreases until it reaches the ground 2.5 seconds after it was hit

Step-by-step explanation:

The given equation for the height of the ball, h(t), in feet, is given as follows

h(t) = 40·t - 16·t²

Therefore;

1) When the ball is hit at 0 seconds, it has a height of h(0) = 40×0 - 16×0² = 0 feet

2) The time to reach maximum height by the ball is found by taking the derivative (slope) of the function for the height of the ball and noting that at maximum height, the derivative (slope) = 0

Therefore;

h'(t) = 40 - 2 × 16·t = 0

40 - 32·t = 0

t = 40/32 = 5/4 seconds = 1.25 seconds

3) The maximum height is then found from the value of the function at t = 5/4 seconds as follows;

h(5/4) = 40×5/4 - 16×(5/4)² = 50 - 25 = 25 feet

4) The time in which the ball reaches the ground where h(t) = 0 is given from the formula for the height of the gulf ball as follows;

h(t) = 0 = 40·t - 16·t²

40·t - 16·t² = 0

(40 - 16·t)·t = 0

Therefore;

t = 0 seconds or t = 40/16 = 2.5 seconds

(40 - 16·t) = 0/t = 0

t = 40/16 = 2.5 seconds

Therefore, the filling the blanks with the correct numbers gives;

When the ball is hit at 0 seconds, it has a height of 0 feet. The ball's height increases until it reaches its maximum height of 25 feet after 1.25 seconds. The ball's height then decreases until it reaches the ground 2.5 seconds after it was hit

4 0

3 years ago

I will mark brainliest to whoever answers most accurately)

Harrizon [31]

Answer:

its the 4th one

Step-by-step explanation:

115000+11000+1200=127200

ans-105500=21700

6 0

4 years ago

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