All categories

3 years ago

In rectangle PQRS, SP = 3 and RS = 4. What is the length of PR? (Hint: use

Mathematics

1 answer:

stich3 [128]3 years ago

7 0

The answer is B). 5.

Both triangles share the same hypotenuse, PR, so you can use the lengths of the legs of the other triangle to find the answer.
4^2 + 3^2 = x^2
25 = x^2
The square root of 25 is 5, so that’s the answer.

You might be interested in

The sum of 4 consecutive number is 330 what is the sum of the largest and the smallest number

Drupady [299]

X + x + 1 + x + 2 + x + 3 = 330
4x + 6 = 330
4x = 324, x = 81
Smallest number: 81
Biggest number: 81 + 3 = 84
Sum: 81 + 84 = 165
The answer is 165

3 0

3 years ago

Help please I really really need help

dem82 [27]

Answer:

In 22 minutes the remaining cell phone battery will be unknown.

In a of minutes the remaining cell phone battery will be 50%.

Step-by-step explanation:

7 0

3 years ago

Insert geometric means in each geometric sequence.

Digiron [165]

Answer:

$\underline{192}, 24, \underline{3}, \underline{\dfrac{3}{8}}, \dfrac{3}{64}$

$\underline{\dfrac{1}8}, \dfrac{1}{4}, \dfrac{1}{2}, \underline{1}$

$81, \underline{27, 9, 3, 1},\dfrac{1}{3}$

Step-by-step explanation:

Given the Geometric sequences:

1. ___, 24, ___, ___, 3/64

2. ___, 1/4, 1/2, ___

3. 81, ___, ___, ___, ___, 1/3

To find:

The values in the blanks of the given geometric sequences.

Solution:

First of all, let us learn about the $n^{th}$ term of a geometric sequence.

$a_n=ar^{n-1}$

Where $a$ is the first term and

$r$ is the common ratio by which each term varies from the previous term.

Considering the first sequence, we are given the second and fifth terms of the sequences.

Applying the above formula:

$ar = 24\\ar^4 = \dfrac{3}{64}$

Solving the above equation:

$r = \dfrac{1}{8}$

Therefore, the sequence is:

$\underline{192}, 24, \underline{3}, \underline{\dfrac{3}{8}}, \dfrac{3}{64}$

Considering the second given sequence:

$ar = \dfrac{1}{4}\\ar^2 = \dfrac{1}{2}\\\text{Solving the above equations}, r = 2$

Therefore, the sequence is:

$\underline{\dfrac{1}8}, \dfrac{1}{4}, \dfrac{1}{2}, \underline{1}$

Considering the third sequence:

$a = 81\\ar^5=\dfrac{1}{3}\\\Rightarrow r = 3$

Therefore, the sequence is:

$81, \underline{27, 9, 3, 1},\dfrac{1}{3}$

5 0

3 years ago

Please solve, -37-2v= -6(2v-3)-5

dezoksy [38]

Answer:

$\huge\boxed{v=5}$

Step-by-step explanation:

$-37-2v=-6(2v-3)-5\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\-37-2v=(-6)(2v)+(-6)(-3)-5\\\\-37-2v=-12v+18-5\\\\-37-2v=-12v+(18-5)\\\\-37-2v=-12v+13\qquad\text{add 37 to both sides}\\\\-37+37-2v=-12v+13+37\\\\-2v=-12v+50\qquad\text{add}\ 12v\ \text{to both sides}\\\\-2v+12v=50\\\\(-2+12)v=50\\\\10v=50\qquad\text{divide both sides by 10}\\\\\dfrac{10v}{10}=\dfrac{50}{10}\\\\v=5$

5 0

3 years ago

Read 2 more answers

X-3 and 2x-1 are factors of 2x^3-px^2-2qx+q. Find the values of p and q.

Nonamiya [84]

Answer:

p =1

q = 9

Step-by-step explanation:

f(x) = 2x³ - px² + 2qx + q

(x - 3) is a factor of f(x)

⇒f(3) = 0

2(3)³ - p*3² - 2q*3 +q = 0

2*27 - 9p - 6q + q = 0

54 - 9p - 5q = 0

-9p - 5q = -54 -------------------(I)

(2x - 1) is a factor of f(x)

2x - 1 = 0

2x = 1

$\st x = \dfrac{1}{2}$

f(1/2) = 0

$2*(\dfrac{1}{2})^{3}-p*(\dfrac{1}{2})^{2}-2q*\dfrac{1}{2}+q=0\\\\2*\dfrac{1}{8}-p*\dfrac{1}{4}-q+q = 0\\\\\dfrac{1}{4}-\dfrac{1}{4}p =0\\\\[Multiply the entire equation by 4]\\\\4*\dfrac{1}{4}-4*\dfrac{1}{4}p=0\\\\$

1 - p = 0

-p = -1

p = 1

Substitute p =1 in equation (I)

-9*1 - 5q = -54

-9 - 5q = -54

-5q = -54 + 9

-5q = -45

q = -45/(-5)

q = 9

8 0

3 years ago