From the equations 7a = 4 and 7a + 4b = 12, one can conclude that b is?Please explain the work behind the solution

Answers

Answer 1

Answer: if 7a=4
then 7a+4b=12 is equal to
4+4b=12
4-4+4b=12-4
4b=8
4b/4=8/4
b=2

Answer 2

Answer: 7a=4
7a+4b=12
you substitute 4 in for 7a because it tells you in the first equation that 7a=4
4+4b=12
next you subtract 4 from each side
4b=8
next you divide 4 from each side
b=2
and theres your answer

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Find all values of k so that the trinomial x^2 + kx - 35 can be factored using integers

Answers

$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$

So the trinomial $x^2+kx-35$ can be factored as long as

$\begin{cases}r_1+r_2=-k\nr_1r_2=-35\end{cases}$

has integer solutions for $r_1,r_2$ . Clearly, both have to be factors of -35, which leaves only a handful of cases:

$(r_1,r_2)=(1,-35)\implies r_1+r_2=-34$
$(r_1,r_2)=(-1,35)\implies r_1+r_2=34$
$(r_1,r_2)=(5,-7)\implies r_1+r_2=-2$
$(r_1,r_2)=(-5,7)\implies r_1+r_2=2$

So the possible values of $k$ are $\pm34$ and $\pm2$ .

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )1) | m | -2 > 0

2) | x - 4 | - 3 > 5

3) | 6 + 9x | ≤ 24

4) | 1 - 5a | > 29

Answers

QUESTION 1

The given inequality is

$|m|-2>0$

We group like terms to get,

$|m|>2$

This implies that,

$-m>2$ or $m>2$ .

We simplify the inequality to get,

$m<-2$ or $m>2$ .

We can write this interval notation to get,

$(-\infty,-2)\cup (2,+\infty)$ .

QUESTION 2

$|x-4|-3\:>\:5$ .

We group like terms to get,

$|x-4|\:>\:5+3$ .

$|x-4|\:>\:8$

We split the absolute value sign to get,

$-(x-4)\:>\:8$ or $x-4\:>\:8$

This implies that,

$x-4\:<\:-8$ or $x-4\:>\:8$

$x\:<\:-8+4$ or $x\:>\:8+4$

$x\:<\:-4$ or $x\:>\:12$

We can write this interval notation to get,

$(-\infty,-4)\cup (12,+\infty)$ .

QUESTION 3

The given inequality is

$|6+9x|\leq 24$

We split the absolute value sign to obtain,

$-(6+9x)\leq 24$ or $(6+9x)\leq 24$

This simplifies to

$6+9x\ge -24$ and $6+9x\leq 24$

$9x\ge -24-6$ and $9x\leq 24-6$

$9x\ge -30$ and $9x\leq 18$

$x\ge -(10)/(3)$ and $x\leq 2$

$-(10)/(3)\leq x\leq2$

We write this in interval form to get,

$[-(10)/(3),2]$

QUESTION 4

The given inequality is

$|1-5a|>29$

We split the absolute value sign to get,

$-(1-5a)>29$ or $1-5a>29$

This simplifies to,

$1-5a\:<\:-29$ or $1-5a\:>\:29$

This implies that,

$-5a\:<\:-29-1$ or $-5a\:>\:29-1$

$-5a\:<\:-30$ or $-5a\:>\:28$

$a\:>\:6$ or $a\:<\:-(28)/(5)$

We write this in interval notation to get,

$(-\infty,-(28)/(5))\cup (6,+\infty)$

(01.01 MC)Why is 3 + (−5) equal to −2? Because it is 5 units to the left of 0 on a horizontal number line Because it is 5 units to the right of 3 on a horizontal number line Because it is 5 units to the left of 3 on a horizontal number line Because it is 5 units to the right of 0 on a horizontal number line

Answers

Answer:

C. Because it is 5 units to the left of 3 on a horizontal number line

Step-by-step explanation:

Directed numbers are numbers which has direction i.e either a negative or positive sign before it. This implies that all numbers are directed, since a number without a sign is taken to be positive.

Thus from the given expression,

3 + (-5) = 3 - 5

= -2

So that,

3 + (-5) = -2

Expressing this on the number line, -2 is 5 units to the left of 3. Thus option C is appropriate.

Still confusing to me!!!!

Answers

Answer:

(a) 2(x -a)

(b) 4x +2h

Step-by-step explanation:

Fill in the function arguments and simplify.

$(f(x)-f(a))/(x-a)=((2x^2-7)-(2a^2-7))/(x-a)\n\n=(2x^2-7-2a^2+7)/(x-a)=(2(x^2-a^2))/(x-a)\n\n=(2(x-a)(x+a))/((x-a))=2(x+a)$

$(f(x+h)-f(x))/(h)=((2(x+h)^2-7)-(2x^2-7))/(h)\n\n=(2x^2+4xh+2h^2-7-2x^2+7)/(h)=(4xh+2h^2)/(h)\n\n=4x+2h$

Polygon JKLM is dilated by a scale factor of 2.5 with point C as the center of dilation, resulting in the image J′K′L′M′. If point C lies on LM¯¯¯¯¯ and the slope of LM¯¯¯¯¯ is 1.75, what can be said about L'M'¯¯¯¯¯¯¯ ?

Answers

I think the slope of L'M' is still 1.75 and it still passes through point C.

Bryan purchased two triangular wall shelves. The sides of each shelf are 8 inches, 10 inches, and 12 inches. He's trying to build a similar, larger triangular shelf to hang between the smaller ones. Which of the dimensions can Bryan use for the larger shelf so that it is similar to the ones he purchased? Justify why the three triangles are similar.A) 4 inches, 5 inches, and 6 inches; All of the side lengths of the smaller triangles have been multiplied by 1/2 which guarantees side-side-side similarity
.
B) 16 inches, 20 inches, and 24 inches; All of the side lengths of the smaller triangles have been multiplied by 2, which guarantees side-side-side similarity.
C) 16 inches, 20 inches, and 24 inches; All of the side lengths of the smaller triangles have been multiplied by 2, which guarantees side-angle-side similarity.
D) 16 inches, 20 inches, and 24 inches; All of the side lengths of the smaller triangles have been multiplied by 2, which guarantees angle-angle-side similarity.

Answers

The correct answer is letter B) 16 inches, 20 inches, and 24 inches; All of the side lengths of the smaller triangles have been multiplied by 2, which guarantees side-side-side similarity. Similar angles shows congruence when their sides are proportional.

Answer:

(B) 16 inches, 20 inches, 24 inches; All the side lengths of the smaller triangles have been multiplied by 2, which guarantees side-side-side similarity.

Step-by-step explanation:

The sides of the triangular wall shelves are given as 8 inches, 10 inches and 12 inches. In order to build a larger, similar triangle he must multiply the sides of the triangle wall by 2 so that it follows the similarity conditions,that is:

$(16)/(8) =(20)/(10)=(24)/(12) =(2)/(1)$

Since, we were given the sides of the triangular wall shelves, therefore the new triangle formed will also be formed of three sides.

Hence, Option (B) is correct in which all the sides of smaller triangles are multiplied by 2 which guarantees side-side-side similarity.

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