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3 years ago

A triangle has a height that is 4 inches shorter than its base. If its area is 30 square inches, find the base and height

Mathematics

1 answer:

klemol [59]3 years ago

8 0

Answer:

120

Step-by-step explanation:

you are multiping you are trying to get the number of the base

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The sides of a square are increased by a scale factor of 6. The perimeter of the smaller square is 10 ft. What is the perimeter

inn [45]

Answer:

60ft

Step-by-step explanation:

I have not done this kind of work in a while but if you scale each side of the square by 6 is should increase to a perimeter of 60ft

4 0

3 years ago

The saucer for Roger's teacup has a radius of 3 centimeters. What is the saucer's circumference?

Fantom [35]

C=dπ and r+r=d so 3+3=6 and 6π≈18.84 so the circumference would be 18.84

7 0

3 years ago

How do you solve -2+6(-6x+6)?

Natali5045456 [20]

Answer: −36x+34

Step-by-step explanation:

Equation: −2+6(−6x+6)

Step 1. distribute the equation:

−2+(6)(−6x)+(6)(6)

−2+−36x+36

Step 2. Finally combine like terms:

−2+−36x+36

(−36x)+(−2+36)

−36x+34

Final Answer:

−36x+34

Hope I could help! :)

4 0

3 years ago

What is negative 13/7 minus negative 5/7 as a fraction

Nimfa-mama [501]

Answer:

-8/7

Step-by-step explanation:

13-7. 8

--------- = --

4 0

3 years ago

Read 2 more answers

Show that if S1 and S2 are subsets of a vector space V such that S1 c S2 then span(S1) c span(S2). In particular, if S1 c S2 the

klemol [59]

Answer:

See proof below

Step-by-step explanation:

Assume that V is a vector space over the field F (take F=R,C if you prefer).

Let $x\in span(S_1)$ . Then, we can write x as a linear combination of elements of s1, that is, there exist $v_1,v_2,\cdots,v_k \in S_1$ and $a_1,a_2,\cdots,a_k\in F$ such that $x=a_1v_1+a_2v_2+\cdots+a_kv_k$ . Now, $S_1\subseteq S_2$ then for all $y\in S_1$ we have that $y\in S_2$ . In particular, taking $y=v_j$ with $j=1,2,\cdots,k$ we have that $v_j\in S_2$ . Then, x is a linear combination of vectors in S2, therefore $x\in span(S_2)$ . We conclude that $span(S_1)\subseteq span(S_2)$ .

If, additionally $S_2\subseteq S_1$ then reversing the roles of S1 and S2 in the previous proof, $span(S_2)\subseteq span(S_1)$ . Then $span(S_1)\subseteq span(S_2)\subseteq span(S_1)$ , therefore $span(S_1)=span(S_2)$ .

5 0

3 years ago