Can someone help me figure this out?

[tex]\\ \sf\longmapsto Interest=Rate\times Time[/tex]

[tex]\\ \sf\longmapsto Interest=5.3(10)[/tex]

[tex]\\ \sf\longmapsto Interest =53[/tex]

It means that if your principal is $100 after 10years you will receive $53 as interest

Step-by-step explanation:

[tex]thank \: you[/tex]

Answer:

12

Step-by-step explanation:

for this you need to draw a straight line and plot the points J,K,L,M,N according to the question.

here,

given K lies between J and M, L is between K and M , M is between K and N.

JN= 30, KM=8

JK=KL=LM

MJ= ?

according to the question,

J--------K-------L--------M-------N

KM= 8

KL+LM= 8 (KL+LM=KM)

KL+KL=8 ( given, LM=KL)

2KL= 8

KL=4

so,

JK= KL=4 and LM= KL= 4

now,

MJ= ?

we know

MJ= JK+KM = 4+8= 12

Answer:

1950

Step-by-step explanation:

because 1950 column has the most highest number which is 11

Answer:

Between 1945 and 1950 ( from 5 to 11 ),

Step-by-step explanation:

Answer:

45

Step-by-step explanation:

CHECK ANSWER:

45  *5 = 225

45+5 = 50

50 < 225

sum < times

I hope this helped !

Answer:

k = 14

Step-by-step explanation:

Prime factorize 32

32 = 2 * 2 * 2 * 2 * 2  = 2⁵

[tex]\frac{2^{12}}{2^{\frac{k}{2}}}= 32\\\\\frac{2^{12}}{2^{\frac{k}{2}}}=2^{5}\\\\2^{12-\frac{k}{2}}=2^{5}[/tex]

Both sides base are same.So, compare exponents

[tex]12-\frac{k}{2}=5\\[/tex]

Subtract 12 from both side

[tex]-\frac{k}{2}=5-12\\\\-\frac{k}{2}=-7\\[/tex]

Multiply both sides by (-2),

[tex](-2)*(-\frac{k}{2})=-7*(-2)\\\\k = 14[/tex]

Answer:

10cm

Step-by-step explanation:

Assuming the shape is a rectangle since it's already stated that it's a parallelogram, and the area is stated, we can use the Pythagorean theorem to find the length

[tex]a^{2} +b^2=c^2[/tex], where c will be the length

isolate c

[tex]c^2=a^2+b^2[/tex]

[tex]c=\sqrt{a^2+b^2}[/tex]

substitute for a and b

[tex]c=\sqrt{8^2+6^2}[/tex]

[tex]c=\sqrt{64+36}[/tex]

[tex]c=\sqrt{100}[/tex]

[tex]c=10[/tex]

Answer:

Infinite solutions

Step-by-step explanation:

x + 4y = 7 --------------(I)

8y = 14 - 2x

⇒ 2x + 8y = 14 -----------(II)

[tex]\frac{a_{1}}{a_{2}}=\frac{1}{2}\\\\\frac{b_{1}}{b_{2}}=\frac{4}{8}=\frac{1}{2}\\\\\frac{c_{1}}{c_{2}}=\frac{7}{14}=\frac{1}{2}\\\\\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}[/tex]

So, infinite solutions

Answer:

G

Step-by-step explanation:

Given the 2 equations

x + 4y = 7 → (1)

8y = 14 - 2x → (2)

make x the subject of (1) by subtracting 4y from both sides

x = 7 - 4y

Substitute x = 7 - 4y into (2)

8y = 14 - 2(7 - 4y) ← distribute

8y = 14 - 14 + 8y

8y = 8y

Since both sides are equal this indicates the system has infinite solutions

[tex]\\ \sf\longmapsto 2x^2+3x^2+3(-1)=5x.x+5x.1[/tex]

[tex]\\ \sf\longmapsto 5x^2-3=5x^2+5x[/tex]

[tex]\\ \sf\longmapsto 5x^2-5x^2-3=5x[/tex]

[tex]\\ \sf\longmapsto 5x=-3[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{-3}{5}[/tex]

Answer:

I found x for you, right there.

Answer:  x is 12 and y is 5

Step-by-step explanation:

Answer:

33/14

Step-by-step explanation:

[tex] - 10 \frac{2}{7} + ( - 4 \frac{4}{11} )[/tex]

[tex] = - \frac{72}{7} \div - ( \frac{48}{11} )[/tex]

[tex] = \frac{72}{7} \times \frac{11}{48} [/tex]

[tex] = \frac{3}{7} \times \frac{11}{2} [/tex]

[tex] = \frac{33}{14} [/tex]

stan dreamcatcher

Answer:

{1, 0, 16}

Step-by-step explanation:

given..

j(x) = x^2-2x+1

put all given values of domain (1,0 and 5 ) in the equation..

the values you get are range of the function

Answer: 1

Step-by-step explanation

0(2)-2(0)+1=0+0+1= 1

1(1)-(2)(1)+1=1-2+1= 1

(5)(2)-(2)(5)+1=10-10+1= 1

Answer:

-117 is irrational number

Answer:

Irrational

Step-by-step explanation:

Irrational number can't be written as a faction, -11pie can't be written as a fraction. Therefore it is a irrational number.

Answer:

O D. 3x2 + 2x-3

Answer:

A.7x2-5x questions 2of 20

Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}\\\\csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{hypotenuse}{adjacent}\\\\cot(\theta)=\frac{adjacent}{opposite}[/tex]

Please note that the terms (opposite) and (adjacent) are relative to the angle uses in the ratio, however the term (hypotenuse) refers to the side opposite the right angle, this side never changes its name. Use these ratios to evaluate the trigonometric functions. Then simplify to prove the identity.

2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

Simplify,

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}[/tex]

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

Simplify,

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

Rationalize the denominator,

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

[tex]2-\sqrt{3}=\frac{(\sqrt{3}-1)^2}{3-1}[/tex]

[tex]2-\sqrt{3}=\frac{3-2\sqrt{3}+1}{2}[/tex]

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

Therefore, this equation is also true.

Answer:

1≤f(x)≤5

Step-by-step explanation:

-1≤x≤1

-2≤2x≤2 (Multiplied by 2 both side)

-2+3≤2x+3≤2+3 (Adding three both sides)

1≤f(x)≤5

the answer is

A. 28, 25, 22

Answer:

A " (1,-2)

B " (4,0)

C " (6,-3)

Step-by-step explanation:

Hope it helped.

° ° °

Step-by-step explanationIf you are reading this say

thank u