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2 years ago

Use algebra tiles to find the difference. Let one yellow tile equal +1 and one red tile equal –1. 12 – (–11)

Mathematics

1 answer:

Anvisha [2.4K]2 years ago

8 0

Answer:

its 23

Step-by-step explanation:

its 12+11 because two negatives make a positive

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Can someone help with the two triangle questions (will mark brainliest!!)

yaroslaw [1]

Answer: Number 1 is c=89 ft

Number 2 is b=3 cm

Step-by-step explanation: a²+b²=c²

For number 1, 39 ft is a and 80 ft is b so

39²+80²=c² so 39² is 39x39=1521 and 80² is 80x80=6400 so 6400+1521=7921

7921=c² and the square root of 7921 is 89 so c=89 ft

For number 2, a=1.6 cm and c=3.4 cm so 3.4²-1.6²=b²

3.4² so 3.4x3.4 which is 11.56

1.6² so 1.6x1.6 which is 2.56

11.56-2.56=c² and 11.56-2.56=9

Square root of 9 is 3 so b=3

So number 2 answer is b=3 cm

I hope this helps

4 0

2 years ago

Everything just the first one, please. Asap, thanks in advance!

alekssr [168]

Question number one is 2/3

7 0

3 years ago

PLEASE HELP ASAP WILL MEDAL BRAINLIEST

Naddik [55]

Given
Present investment, P = 22000
APR, r = 0.0525
compounding time = 10 years
Future amount, A

A. compounded annually
n=10*1=10
i=r=0.0525
A=P(1+i)^n
=22000(1+0.0525)^10
=36698.11

B. compounded quarterly
n=10*4=40
i=r/4=0.0525/4
A=P(1+i)^n
=22000*(1+0.0525/4)^40
=37063.29

Therefore, by compounding quarterly, she will get, at the end of 10 years investment, an additional amount of
37063.29-36698.11
=$365.18

5 0

2 years ago

Country days scholarship rounds receive a gift of $135000. The money is invested in stock, bonds, and CDs. CDs pay 2.75% interes

Vsevolod [243]

Answer:

CDs — $10,000
bonds — $80,000
stocks — $45,000

Step-by-step explanation:

Let the variables c, b, s represent the dollar amounts invested in CDs, stocks, and bonds, respectively. Then the problem statement gives us 3 relations between these 3 variables:

c + b + s = 135000 . . . . . . . . . . . . . . . . . total invested

0.0275c +.045b +0.104s = 8555 . . . . . total income earned

-c + b = 70000 . . . . . . . . . . . . . . . . . . . . . 70,000 more was in bonds than CDs

Using the third equation to write an expression for b, we can substitute into the other two equations.

b = 70000 +c . . . . . . . . . . . . . . . . expression we can substitute for b

c + (70000 +c) +s = 135000 . . . . substitute for b in the first equation

2c +s = 65000 . . . . . . . . . . . . . . . . [eq4] simplify

.0275c +.045(70000 +c) +.104s = 8555 . . . . . substitute for b in 2nd eqn

.0725c +.104s = 5405 . . . . . . . . . . [eq5] simplify

Using [eq4], we can write an expression for s that can be substituted into [eq5].

s = 65000 -2c . . . . . . . expression we can substitute for s

0.0725c +0.104(65000 -2c) = 5405

-0.1355c = -1355 . . . . . . . . . . . . . . . . . . . . subtract 6760, simplify

c = 1355/.1355 = 10,000

s = 65000 -2×10000 = 45,000

b = 70000 +10000 = 80,000

The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

_____

Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

$\left[\begin{array}{ccc|c}1&1&1&135000\\0.0275&0.045&0.104&8555\\-1&1&0&70000\end{array}\right]$

6 0

3 years ago

I need the answer for this exact question

denis23 [38]

Answer:

$y=\frac{7}{2}x-20$

Step-by-step explanation:

Let the equation of the line be $y-y_1=m(x-x_1)$ where, 'm' is its slope and $(x_1,y_1)$ is a point on it.

Given:

The equation of a known line is:

$y=-\frac{2}{7}x+9$

A point on the unknown line is:

$(x_1,y_1)=(4,-6)$

Both the lines are perpendicular to each other.

Now, the slope of the known line is given by the coefficient of 'x'. Therefore, the slope of the known line is $m_1=-\frac{2}{7}$

When two lines are perpendicular, the product of their slopes is equal to -1.

Therefore,

$m\cdot m_1=-1\\m=-\frac{1}{m_1}\\m=-\frac{1}{\frac{-2}{7} }=\frac{7}{2}$

Therefore, the equation of the unknown line is determined by plugging in all the given values. This gives,

$y-(-6))=\frac{7}{2}(x-4)\\y+6=\frac{7}{2}x-14\\y=\frac{7}{2}x-14-6\\\\y=\frac{7}{2}x-20$

The equation of a line perpendicular to the given line and passing through (4, -6) is $y=\frac{7}{2}x-20$ .

3 0

2 years ago