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2 years ago

A triangle has sides measuring 5 inches and 8 inches. If x represents the

Mathematics

2 answers:

lord [1]2 years ago

6 0

Answer:

For me,the answer will be A.

im not sure

correct me if im wrong

love ya

frez [133]2 years ago

6 0

Answer:

Step-by-step explanation:

Given 2 sides of a triangle then the third side x is in the range

difference of 2 sides < x < sum of 2 sides , that is

8 - 5 < x < 8 + 5 , then

3 < x < 13 → B

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1. What is factorization? <br>2.Find the factorization of 891 and please explain!<br><br>Thank you!

Anastaziya [24]

Answer:

In mathematics, factorization (also factorisation in some forms of British English) or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.Prime factorization of 891 = 1×3×3×3×3×11= 1×34×11

Step-by-step explanation:

1. We write number 891 above a 2-column table

2. We divide 891 by the smallest possible prime factor

3. We write down on the left side of the table the prime factor and next number to factorize on the ride side

4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors)

5. We continue until we reach 1

6 0

2 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

2 years ago

Solve triangle ABC with m

Irina18 [472]

Yeah I’ll get right on that.

3 0

2 years ago

4(x-11)^2=25

netineya [11]

Expand the square: 4(x-11)^ =25
4(x-11)(x-11)=25

Distribute : 4(x-11)(x-11)=25
4(x(x-11)-11(x-11))=25

Distribute: 4(x(x-11)-11(x-11))=25
4(x^-11x-11(x-11))=25

Solution: x=17/2
x=27/2

hope that helps

8 0

2 years ago

The $25.00 shirt costs the store $20.00. Find the markup rate.

pochemuha

Answer:

25%

Step-by-step explanation:

3 0

3 years ago