All categories

3 years ago

I NEED HELP QUICK! PLEASE

Mathematics

1 answer:

blagie [28]3 years ago

7 0

Answer:

(a, 0), (0, b), (-a, 0), (0, - b)

Step-by-step explanation:

Since, all vertices of the rhombus lie in the x and y axes.

x - axis is given by a and y - axis is given by b.

Therefore, coordinates of vertices lying on x - axes will be (a, 0) & (-a, 0) and coordinates of vertices lying on y - axes will be (0, b) & (0, - b).

You might be interested in

Ok this one is pearson realize if u know this help

makvit [3.9K]

There are 12 student(s) in the group, because there is 12 one-third part(s) in 4 while jars

7 0

2 years ago

Read 2 more answers

Simplify 3^-3<br> Show work

dmitriy555 [2]

$\bf a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 3^{-3}\implies \cfrac{1}{3^3}\implies \cfrac{1}{27}$

5 0

3 years ago

Help please . 10 points

V125BC [204]

Answer:

xtsLztfahlzkfxsh bldg MS much k stick us rich l dips ssssseereeeeeexxxxxxxx

3 0

3 years ago

A regular hexagon is shown below. Find the value of x.<br> (11x + 21)°

noname [10]

Answer:

x=9

Step-by-step explanation:

The expression is for the interior angle of the hexagon; one interior angle is equal to .

Since an interior angle of a hexagon measures 120°, we have the equality

Now it is just a matter of solving for

11x=120-21

11x=99

x=99/11

x=9

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