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4 years ago

Whats the answer to this question

Mathematics

2 answers:

Misha Larkins [42]4 years ago

7 0

Yes because 10-23 is -13 and -10 is greater than that

andriy [413]4 years ago

4 0

Answer:

Yes.

Step-by-step explanation:

We have been given an inequality. We are asked to determine whether 10 is a solution for the given inequality or not.

$x-23\leq -10$

Upon substituting $x=10$ in our given inequality, we will get:

$10-23\leq -10$

$-13\leq -10$

Since the solution $x=10$ satisfies our given inequality, therefore, 10 is a solution for the given inequality.

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(1 point)

juin [17]

Answer:

Step-by-step explanation:

2 times 150

7 0

3 years ago

Please answer this!!

balu736 [363]

Answer:

3,724 cubic inches

Step-by-step explanation:

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5 0

3 years ago

Simplify 270 ÷ 3[(4 - 3)3 - (-9)] - 53

Andreyy89

270 / 3 ( (4 - 3) 3 - (-9) ) - 53
270 / 3 (1 x 3 - (-9) ) - 53
270 / 3 (3 + 9) - 53
270 / 3 x 12 - 53
In PEMDAS, with multiplication and division, you must go left to right, not multiplication then division.
90 x 12 - 53
1080 - 53
1027 is your answer

4 0

3 years ago

Read 2 more answers

Y=6x-11 -2x-3y=-7 solve each system of equation by substitution

Anit [1.1K]

Answer:

(2, 1 )

Step-by-step explanation:

Given the 2 equations

y = 6x - 11 → (1)

- 2x - 3y = - 7 → (2)

Substitute y = 6x - 11 into (2)

- 2x - 3(6x - 11) = - 7 ← distribute and simplify left side

- 2x - 18x + 33 = - 7

- 20x + 33 = - 7 ( subtract 33 from both sides )

- 20x = - 40 ( divide both sides by - 20 )

x = 2

Substitute x = 2 into (1) for corresponding value of y

y = 6(2) - 11 = 12 - 11 = 1

Solution is (2, 1 )

5 0

3 years ago

In this exercise, T: R2 → R2 is a function. For each of the following parts, state why T is not linear. (a) T(a1, a2)= (1, a2) (

DENIUS [597]

Answer:

a) T is not homogeneous
b) T is not additive
c) T is not homogeneous
d) T is not additive
e) T is not additive

Step-by-step explanation:

In each case there is an example where one property of linear functions fails.

a) $T(2(1,0))=T((2,0))=(1,0); 2T((1,0))=2(1,0)=(2,0)$ . These vectors are not equal, then T doesn't satisfy the condition of scalar multiplication (homogeneity).

b) $T((1,2)+(2,3))=T(3,5)=(3,9); T((1,2))+T((2,3))=(1,1)+(2,4)=(3,5)$ . Because these vectors are not equal, T doesn't satisfy the property of vector addition (additivity).

c) $T(\frac{1}{2}(\pi,0))=T((\frac{\pi}{2},0))=(\sin(\frac{\pi}{2}),0)=(1,0); \frac{1}{2}T((\pi,0))=\frac{1}{2}(0,0)=(0,0)$ so T is not homogeneous.

d) $T((-1,0)+(1,0))=T(0,0)=(0,0); T((-1,0))+T((1,0))=(1,0)+(1,0)=(2,0)$ then T is not additive.

e) $T((0,0)+(1,0))=T(1,0)=(2,0); T((0,0))+T((1,0))=(1,0)+(2,0)=(3,0)$ then T is not additive.

8 0

3 years ago