Whats the aream of this rhombus

A football team lost 5 yards on one play and gained 12 yards on the next play. Find the overall change in field position.

MrRissso [65]

Answer:

Equation: -5 + 12

Overall change: 7 yards

Step-by-step explanation:

In the problem, it says "lost 5 yards". "Lost" is another way of saying subtraction or negative. Therefore, our starting number is -5.

The problem also says "gained 12 yards". "Gained" is another was of saying addition or positive. Therefore, our second number is positive 12.

Together, these two numbers look like this: -5 + 12. To add them, go 12 to the right of -5 if on a number line. That would be like this: -5, -4, -3, -2, -1, 0, etc until you get to 7 which should be 12 more than -5.

Therefore, the answer is 7 yards.

6 0

3 years ago

For what value(s) of k will the relation not be a function?

Roman55 [17]

Answer:

b=k3+24k2−3700k

r=(k3−5k2+3k−5)((−k)(4))

87(|k+1|)+2088

Step-by-step explanation:

4 0

3 years ago

I will give brainlist!

Ilya [14]

Answer:

The length of x is 4

Step-by-step explanation:

The scale factor is Large/small where 10/8 which is DF/AC.

THE SCALE factor is 5/4.

To find the length of X is in AB we use the scale factor 5/4 and the corresponding DE.

5/4 = DE/AB

5/4 = 5/X

X = 4

To find Y in EF will use the corresponding BC which is 6.

5/4 = EF/BC

5/4 = Y/6

Y= 7.5

info: use CORRESPONDING ABC AND DEF .

3 0

3 years ago

A company produces a certain product, and each unit of this product may have 3 different types of defects. Let Di, D2,Ds represe

Stella [2.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) 0.88

b) 0.02

c) 0.01

d) 0.99

Step-by-step explanation:

Step one: State the given parameters

$P(D_{1} ) = 0.12$ $P(D_{2} ) = 0.07$

$P(D_{3} ) = 0.05$ $P (D_{1} U D_{2} ) = 0.13$

$P(D_{1}n D_{2}n D_{3}) = 0.01$ $P(D_{1} U D_{3}) = 0.14$

Step 2 : Obtain the probability that a unit does not have a type 1 defect

$P(\frac{}{D_{1} })$ = $1 -P(D_{1} )$

= $1 - 0.12$

= 0.88

Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?

The probability of the unit having both type 2 and type 3 defect is denoted as $P(D_{2} n D_{3} )$

This is calculated as

$P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\ = 0.07 + 0,05 - 0.13$

= 0.02

Therefore P(D_{2} n D_{3} ) = 0.02

Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect

Let $P(\frac{}{D_{1}} n D_{2} n D_{3} )$ denote the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.

This can be calculated as follows :

$P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})$

= 0.02 - 0.01

= 0.01

Step 4 : Obtain the probability that a unit has at most two defects

P(at most 2 defects) = 1 - P(all three defects)

= $1- P(D_{1} n D_{2}nD_{3})$

= 1 - 0.01

= 0.99

7 0

3 years ago

A simulation is done to represent kicking 5 field goals in a single game with a 72% probability of making each one. A 1 represen

Blababa [14]

Answer:

You are correct that the cross section is a rectangle, because a rectangular pyramid has 4 sides connected to the top of the rectangle.

Step-by-step explanation:

8 0

3 years ago