Ask question

Ask question

All categories

3 years ago

13

jewels score was 5 over par on the first 1 holes. her score was 4 under on the second 9 holes. what is this but in integers

1 answer:

melamori03 [73]3 years ago

7 0

Answer:

1

Step-by-step explanation:

In golf, if you are over par, then your score is represented by values over zero and with a plus. Since Jewels' score was 5 over on the first hole (dang... that's a quintuple bogey (which is horrible)), she had a +5 on the first hole. But for the next 9 holes, she scored 4 under, which is represented by -4. Add 5 and -4 and you get +1. She still has 8 holes left in that round if she wants to get below par.

You might be interested in

Adam, Matt, and Clinton sold sandwiches in the ratio of 2:7:3 to raise money their basketball team. Matt sold 91 sandwiches. How

zavuch27 [327]

Answer:

156

Step-by-step explanation:

Let,

Sandwiches sold by Adam, Matt and Clinton = 2x, 7x and 3x

Total sandwiches = 2x + 7x + 3x

=> 12x

But its given that Matt sold 91 sandwiches

So,

7x = 91

x = 91/7

x = 13

=> Total sandwiches = 12x = 12(13) = 156

5 0

2 years ago

100 POINTS IF U GET THIS RIGHT!

Darina [25.2K]

Answer7.00000

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

1.- What is 'a' for this hyperbola?

SpyIntel [72]

Answer:

The standard form of a hyperbola with vertices and foci on the x-axis:

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

where:

center: (h, k)
vertices: (h+a, k) and (h-a,k)
Foci: (h+c, k) and (h-c, k) where the value of c is c² = a² + b²
Slopes of asymptotes: $\pm\left(\dfrac{b}{a}\right)$

<h3><u>Part 1</u></h3>

The center of the given hyperbola is (0, 0), therefore:

$\implies \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

Therefore $(\pm a,0)$ are the vertices. From inspection of the graph, $a=2$ .

<h3><u>Part 2</u></h3>

Choose two points on the asymptote with the positive slope:

(0, 0) and (4, 6)

Use the slope formula to find the slope:

$\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{6-0}{4-0}=\dfrac{3}{2}$

<h3><u>Part 3</u></h3>

Use the <u>slopes of asymptotes</u> formula, compare with the slope found in part 2:

$\implies \dfrac{b}{a}=\dfrac{3}{2}$

Therefore, $b=3$

<h3><u>Part 4</u></h3>

Substitute the found values of $a$ and $b$ into the equation from part 1:

$\implies \dfrac{x^2}{2^2}-\dfrac{y^2}{3^2}=1$

$\implies \dfrac{x^2}{4}-\dfrac{y^2}{9}=1$

4 0

1 year ago

Simplify (2p+3)2 - (2p-3)2

gayaneshka [121]

(2p + 3)2 - (2p - 3)2 = 4p+6-4p+6=12

8 0

2 years ago

Can a person train to become better at holding their breath? An experiment was designed to find out. A group of 12 volunteers we

mixer [17]

Answer:

GROUP 1

Step-by-step explanation:

8 0

2 years ago