Do the following side lengths form a triangle? 28, 50, 22

Write each ratio as a fraction in the lowest terms

stich3 [128]

Answer:

1/4

Step-by-step explanation:

$(4\times \frac{3}{8} ) : (12\times \frac{1}{2} )\\\\\frac{4\times \frac{3}{8}}{12\times \frac{1}{2}}\\\\\mathrm{Multiply\:}12\times \frac{1}{2}\::\quad 6\\=\frac{4\times \frac{3}{8}}{6}\\\\\mathrm{Multiply\:}4\times \frac{3}{8}\::\quad \frac{3}{2}\\=\frac{\frac{3}{2}}{6}\\\\=\frac{3}{2\times \:6}\\\\=\frac{3}{12}\\\\=\frac{1}{4}$

4 0

3 years ago

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Find two consecutive odd integers such that 53 more than the lesser is four times the greater

viva [34]

Let x and x+2 the 2 odd consecutive numbers. Now translate from English language to Math's:
x+53 = 4(x + 2), Expand :
x+53 = 4x + 8
53 - 8 = 4x - x

45 = 3x and x = 15. So the 1st is 15 and the 2nd is 17

4 0

4 years ago

Find the value of the expression. c²b for b = 5 and c = 3

kicyunya [14]

Answer:

45

Step-by-step explanation:

c^2*b

We know that c is 3, and b is 5, so we can substitute them in

3^2*5

Solve the exponent first

9*5

Multiply

45

7 0

3 years ago

Read 2 more answers

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .

5 0

4 years ago

7. Find the measure of <1. is 89, measur 105

Kazeer [188]

Answer:

where is the diagram? please provide a diagram.

5 0

3 years ago