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GarryVolchara [31]
3 years ago
5

An isosceles triangle has a perimeter of 36 inches. Its base is 2

Mathematics
1 answer:
astra-53 [7]3 years ago
3 0

The lengths of the sides of the triangle are 8, 8, 20

Explanation:

Given that the perimeter of an isosceles triangle is 36 inches.

The base of the triangle is 2\frac{1}{2} times longer than each of its legs.

We need to determine the lengths of the sides of the triangle.

<u>Lengths of the sides:</u>

Let x denote the lengths of the sides of the triangle.

The base of the triangle is given by

2\frac{1}{2}x=\frac{5}{2}x=2.5x

Perimeter of the isosceles triangle = Sum of the three sides of the triangle.

Thus, we have,

36=x+x+2.5x

36=4.5x

8=x

Thus, the length of the sides of the isosceles triangle is 8 inches.

Base of the triangle = 2.5(8)=20 \ inches

Hence, the three sides of the isosceles triangle are 8, 8, 20

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Find the output, kkk, when the input, ttt, is -7−7minus, 7.<br> k = 10t-19k=10t−19
Sati [7]

The output k at input t = -7 is -89

Step-by-step explanation:

In order to find the value of a expression on given number, we put the number in given expression in place of independent variable

So,

Given expression is:

k = 10t-19

It is given that the value of t has to be -7

so putting t=-7 in the expression

k = 10(-7)-19\\k = -70-19\\k = -89

Hence,

The output k at input t = -7 is -89

Keywords: Expressions, input, output

Learn more about expressions at:

  • brainly.com/question/9723881
  • brainly.com/question/9720317

#LearnwithBrainly

8 0
3 years ago
Read 2 more answers
a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space
agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are ifv = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V

Property three does now no longer follow: Suppose that Property three is legitimate, shall we namev = a * x ^ 2 +bx +cthe neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently 0 = O + v = (O  x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= 0 If O is the neuter, then it ought to restore x², but 0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^  x^ 2 +(vl^ × wl)^  x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 thenv = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= zero If O is the neuter, then it ought to restore x², but zero + x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could usez = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)

Read more about polynomials :

brainly.com/question/2833285

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8 0
11 months ago
4 loads of stone weigh 2/3 ton. Find the weight of 1 load of stone.
Vladimir79 [104]

To find the weight of 1 load of stone, divide 2/3 by 4.

To divide fractions, use keep change flip.

Keep the first fraction the same, change the division sign into a multiplication sign, and flip the second fraction.

I attached the work and answer for this problem. If you have any questions please feel free to ask.

Hope this helps.

7 0
3 years ago
Decide Whether each statement is true or false:
vazorg [7]
If they’re about coordinates, the first one is true and the next two are not.
But if it’s not about coordinating they’re all true.
8 0
2 years ago
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FIRST ANSWER GET BRAINLEST
liraira [26]

Let's call x the price of a box of slime.

Gianna spent 4 × $58 for art supplies so 4(58) + 8x all together.  We have

392 ≤  4(58) + 8x ≤ 472

We can work with both inequalities at once.  4(58)=232 so we subtract that from all three sides.

160 ≤ 8x ≤ 240

Divide all three sides by 8, a positive number so the inequalities don't flip.

20 ≤ x ≤ 30

Answer: Least: $20, Most: $30


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3 years ago
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