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2 years ago

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Which product is negative? (Negative StartFraction 3 over 8 EndFraction) (Negative StartFraction 5 over 7 EndFraction) (one-four

th) (StartFraction 3 over 8 EndFraction) (Negative StartFraction 5 over 7 EndFraction) (Negative one-fourth) (StartFraction 3 over 8 EndFraction) (StartFraction 5 over 7 EndFraction) (one-fourth) (Negative StartFraction 3 over 8 EndFraction) (Negative StartFraction 5 over 7 EndFraction) (negative one-fourth)

2 answers:

Dovator [93]2 years ago

5 0

answer:

Negative 2 and 1 over 8

Step-by-step explanation:

nasty-shy [4]2 years ago

4 0

Answer:

$(D)\left(-\dfrac38\right)\left(-\dfrac57\right)\left(-\dfrac14\right)$

Step-by-step explanation:

The given options are:

$(A)\left(-\dfrac38\right)\left(-\dfrac57\right)\left(\dfrac14\right)\\(B)\left(\dfrac38\right)\left(-\dfrac57\right)\left(-\dfrac14\right)\\(C)\left(\dfrac38\right)\left(\dfrac57\right)\left(\dfrac14\right)\\(D)\left(-\dfrac38\right)\left(-\dfrac57\right)\left(-\dfrac14\right)$

The key to determining which product is negative is to understand the rule of sign multiplication.

Now:

The product of even negative terms is positive
The product of odd negative terms is negative.
The product of positive will always be positive.

In Options A and B, the number of negative signs is even, therefore our result is positive.

In option C, all the terms are positive, therefore our result will be positive.

In Option D, the number of negative signs is odd, therefore our result is negative.

You might be interested in

Find two complex numbers that have a sum of i10 a different of -4and a product of -29

lianna [129]

<h2>-2+5i and 2+5i</h2>

Step-by-step explanation:

Let the complex numbers be $a+ib\textrm{ and }c+id$ .

Given, sum is $10i$ , difference is $-4$ and product is $-29$ .

$(a+c)+i(b+d)=10i$ ⇒ $a+c=0,b+d=10$

$(a-c)+i(b-d)=-4$ ⇒ $a-c=-4,b-d=0$

$a=-2,c=2,b=5,d=5$

$(a+ib)(c+id)=(-2+5i)(2+5i)=-4-25=-29$

Hence, all three equations are consistent yielding the complex numbers $-2+5i\textrm{ and }2+5i$ .

8 0

2 years ago

A 4/7 inch pipe is to be shortened to 3/8 inch. How much must be removed?

11Alexandr11 [23.1K]

11/56 inches or 0.196 inches must be removed.

The pipe measures 4/7 inches but needs to be reduced to 3/8 inches.

In order to find out the inches to be removed, you must subtract the length that the pipe should be from the length that it currently is.

<em>Length to be removed = 4/7 - 3/8</em>

You need a common denominator so find the lowest common factor of both denominators:

= 56

In the shared fraction, multiply the numerator by the number you get when you divide 56 by the denominator.

= 56/7 = 8 8 x 4 = 32

= 56 / 8 = 7 7 x 3 = 21

= (32 - 21) / 56

= 11 / 56 inches

= 0.196 inches

In conclusion, 11/56 inches must be removed to get the pipe to 3/8 inches

<em>Find out more at brainly.com/question/4681199.</em>

6 0

1 year ago

Jessica rides the bus 8and 4/5 miles each day. WHATS is The total Numbers of miles she rindes in 21 day's

lesantik [10]

201 miles i believe, hope this helps!

5 0

2 years ago

Read 2 more answers

Use the binomial theorem to show that 7^n + 2 is divisible by 3, where n is a positive integer.

masha68 [24]

By the binomial theorem,

$7^n=(6+1)^n=6^n+n6^{n-1}+\dbinom n26^{n-2}+\cdots+\dbinom n{n-2}6^2+n6^1+1$

Each term containing a factor 6 is divisible by 3, since 3 divides 6. The remaining 1 and 2 sum to 3, which is also divisible by 3.

7 0

1 year ago

Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.

xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form $p(t)=at^2$ , where a is in ℝ.

2.Let W2 be the set of all polynomials of the form $p(t)=t^2+a$ , where a is in ℝ.

3. Let W3 be the set of all polynomials of the form $p(t)=at^2+at$ , where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

$p_1 (t) = at^2, p_2(t)=bt^2$ .

We must check that p_1+p_2(t) is in W1.

Note that

$p_1(t)+p_2(t) = at^2+bt^2 = (a+b)t^2$

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say $p_1(t) = a(t^2+t), p_2(t)= b(t^2+t)$ . Then

$p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)$

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 $p(t) = at^2$ and a real scalar b. Then

$bp(t) = b(at^2) = (ba)t^2)$ .

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 $p(t) = a(t^2+t)$ and a real scalar b. Then

$bp(t) = b(a(t^2+t)) = (ba)(t^2+t)$ .

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.

6 0

2 years ago