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zloy xaker [14]

3 years ago

10

Last Thursday you bought a total of 33 school supply items. You spent $58 Pencils are $1.50 and glue sticks are $2 each. How man

y of each item did you buy? (5 pts)

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer:

16 pencils and 17 glue sticks

Step-by-step explanation:

X=pencils

Y=glue sticks

X+Y=33

1.50X+2Y=58

times ten to get whole numbers

=15X+20Y=580

same for e1

=10X+10Y=330

15X+20Y=580

10X+10Y=330

elimination mthd

150X+200Y=5800

150X+150Y=4950

50Y= 850

<u>Y</u><u>=</u><u>1</u><u>7</u>

<u>X</u><u>+</u><u>17</u><u>=</u><u>3</u><u>3</u>

X=33-17

X=16

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What expression is equivalent to (9x^2 + 2x -7)(x - 4)

Dima020 [189]

Answer: C

Step-by-step explanation:

We can use the expanding rule to get that one if the expressions is

(9x^2 + 2x - 7)x + (9x^2 + 2x - 7)(-4)

3 0

2 years ago

Which equation could be used to solve the following system?

nikdorinn [45]

we have

$x+y=11$

$x=11-y$ --------> equation A

$4x^{2} -3y^{2}=8$ ----> equation B

Substitute equation A in equation B

$4(11-y)^{2} -3y^{2}=8$

therefore

<u>the answer is the option B</u>

$4(11-y)^{2} -3y^{2}=8$

8 0

3 years ago

Read 2 more answers

In the following system of equations, what would you do to eliminate x? 2x + 4y = 10 x - 3y = 12

Anvisha [2.4K]

2x +4y =10x-3y =12

2x +4y =12 -----eq. 1

10x-3y =12 -------eq.2

2x =12 -4y [using eq. 1]
x = (12-4y)/ 2
= 6 -2y

Now, substitute the value of x in eq. 2
10 (6-2y)-3y =10
60 -20y -3y =10
-23 y = 10 -60
y = -50 / -23
y = 50/23

Now, substitute this value of y in any equations given above

10 x-3 (50/23) =12
10x - (150/23)= 12
x= 426/230

8 0

4 years ago

I need answer and specific explanation plz...

vitfil [10]

Answer:

The smallest model - bottom right - in the question diagram represents $\sqrt[3]{64}=4$ .

Step-by-step explanation:

Considering the radical expression

$\sqrt[3]{64}$

Lets simply this radical expression first

As

$\sqrt[3]{64}$

$\mathrm{Factor\:the\:number:\:}\:64=4^3$

$=\sqrt[3]{4^3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a,\:\quad \:a\ge 0$

$\sqrt[3]{4^3}=4$

$=4$

Therefore, $\sqrt[3]{64}=4$

Now, as we can determine that $\sqrt[3]{64}=4$ . So, the smallest model in the question diagram represents $\sqrt[3]{64}=4$ as each face of the cube of the smallest model in the diagram - bottom right - has 4 squares.

Therefore, the smallest model - bottom right - in the question diagram represents $\sqrt[3]{64}=4$ .

Keywords: square cube root, radical expression

Learn more about radical expression from brainly.com/question/13984232

#learnwithBrainly

7 0

3 years ago

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago