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3 years ago

Write the equation for the function graphed above g(x)=

Mathematics

1 answer:

Svetradugi [14.3K]3 years ago

6 0

-(x+1)^2 + 3

move 3 units up, 1 unit to the left, and a reflection from x^2.

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The width of a rectangle is 3 inches less than twice the length. If the length of the rectangle is represented by L, write an al

Klio2033 [76]

2L - 3 = Width
so if Width = w
w = 2L - 3

7 0

3 years ago

Find the area of the following shape,

lord [1]

$\bold{ANSWER:}$
125.71 cm^2

$\bold{SOLUTION:}$

5 0

2 years ago

Keith started saving for retirement at age 45 with plans to retire at age 70. He invested an average of $500 per month in variou

saveliy_v [14]

500×((1+0.06÷12)^(12×25)−1)÷(0.06÷12)
=346,496.98

8 0

3 years ago

In a certain lottery, you must select 5 numbers (in any order) out of 26 correctly to win. a. How many ways can 5 numbers be

Step2247 [10]

Answer:

a. 7893600

$b.\ \dfrac{1}{26}$

Step-by-step explanation:

Given that there are 26 numbers out of which 5 numbers are to be chosen.

Here repetition is not allowed.

For each of the 5 cases, the total number of possibilities will keep on decreasing by 1.

1st case, number of possibilities = 26

2nd case, number of possibilities = 25

3rd case, number of possibilities = 24

4th case, number of possibilities = 23

5th case, number of possibilities = 22

a. Total number of possibilities = 26 $\times$ 25 $\times$ 24 $\times$ 23 $\times$ 22 = <em>7893600</em>

b. Probability of winning by choosing one number:

Formula for probability of an event E can be observed as: $P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

Number of favorable cases = 1

Total number of cases = 26

So, required probability:

$P(E) = \dfrac{1}{26}$

So, the answers are:

a. 7893600

$b.\ \dfrac{1}{26}$

5 0

3 years ago

Triangles ΔABC ≅ ΔBAD so that C and D lie in the opposite semi-planes of segment AB. Prove that segment CD bisects segment AD.

kolbaska11 [484]

Answer:

See explanation

Step-by-step explanation:

Triangles ΔABC and ΔBAD are congruent. So,

AB ≅ BA;
AC ≅ BD;
BC ≅ AD;
∠ABC ≅ ∠BAD;
∠BCA ≅ ∠ADB;
∠CAB ≅ ∠DBA.

Consider triangles AEC and BED. In these triangles,

AC ≅ BD;
∠EAC ≅ ∠EBD (because ∠CBA ≅ ∠BAD);
∠AEC ≅ ∠BED (as vertical angles).

So, ΔAEC ≅ ΔBED. Thus,

AE ≅ EB.

This means that segment CD bisects segment AD.

6 0

3 years ago