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PSYCHO15rus [73]

3 years ago

12

PLEASE HELP IM DUMB AND I NED THIS ASAP

1 answer:

netineya [11]3 years ago

6 0

Width (7.8) x length (7.8) x width (7.8) = 474.552

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If h(x)=1-2/3x, find h(-6).

Alex

Answer:

5

Step-by-step explanation:

Put -6 where x is and do the arithmetic.

$h(x)=1-\dfrac{2}{3}x\\\\h(-6)=1-\dfrac{2}{3}(-6)=1-\dfrac{-12}{3}=1-(-4)=1+4\\\\\boxed{h(-6)=5}$

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s2008m [1.1K]

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0+0=0

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irina1246 [14]

4/3 or 1.3. hope this helps.

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3 years ago

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Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

2 years ago