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san4es73 [151]
1 year ago
5

Kyla has the number of cherries that Denise has. If Kyla has 46 cherries, how many cherries do they have altogether?​

Mathematics
2 answers:
pogonyaev1 year ago
5 0

Answer:

92

Step-by-step explanation:

46 + 46 = 92

Marrrta [24]1 year ago
5 0
Answer: 92
Explanation:
46+46=92
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I just need the answer to check and see if i got mine right.
Rama09 [41]
Yes you are correct.........
5 0
2 years ago
5.146 rounded to the nearest tenth
Gnom [1K]

Answer:

5.1

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Which value of m will create a system of parallel lines with no solution? y=mx-6 8x-4y=12 A coordinate grid with one line labele
nydimaria [60]

Answer:

A system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2

Step-by-step explanation:

The equation of the given line is 8·x - 4·y = 12

Which gives;

8·x- 12= 4·y

y = 2·x - 3

Given that the line passes through the points (0, -3) and (1, -1), we have;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

When (x₁, y₁) = (0. -3) and (x₂, y₂) = (1, -1), we have;

Slope, \, m =\dfrac{(-1)-(-3)}{1-(0)} = 2

y - (-3) = 2×(x - 0)

y = 2·x - 3 which is the equation of the given line

For the lines 8·x - 4·y = 12, which is the sane as y = 2·x - 3 and the line y = m·x  - 6 to have no solution, the slope of the two lines should be equal that is m = 2

Given that the line passes through the point (1.5, 0), we have;

y - 0 = 2×(x - 1.5)  

y = 2·x - 3...................(1)

For the equation, y = m·x  - 6, when m = 2, we have;

y = 2·x  - 6..................(2)

Solving equations (1) and (2) gives;

2·x - 3 = 2·x  - 6, which gives;

2·x - 2·x=  - 3 - 6

0 = 9

Therefore, a system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2.

4 0
3 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
Tools are us rents leaf blowers for a base price of $24 +9 dollars per hour XYZ rentals rent leaf blowers for a base price of $1
Sidana [21]

Price per hour of first case :

P_1 = 24 + 9a ( Here, a is number of hours )

Price per hour of second case :

P_2 = 18 + 11b ( Here, b is number of hours )

Let, after x hours both company charges the same.

P_1=P_2\\\\24+9x = 18 + 11x\\\\2x=6\\\\x = 3\ hours

Hence, this is the required solution.

6 0
3 years ago
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