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2 years ago

4. What are the coordinates of the x-intercept for the equation y= -1x + 3?

Mathematics

1 answer:

uysha [10]2 years ago

5 0

Answer:

Step-by-step explanation:

The x intercept is the value of x when y = 0. Use the equation to find the value of x:

y= -1x + 3

0 = -1x + 3

x = 3

The point is (3,0)

You can also graph it. Attached.

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Please help! <br><br> 6.2 divided by (2x + 1)= 2 1/3 divided by 4

guajiro [1.7K]

Answer:

x=2/3

Step-by-step explanation:

(2x+1)/2 = 2 1/3/4

Multiply 4 on both sides:

2x+1=2 1/3

Rewrite:

2x+1=7/3

Subtract 1 on both sides:

2x=4/3

Divide 2 on both sides:

x=2/3

Thus x=2/3

Hope this helps!

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2 years ago

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Which of these sets could represent the side lengths of a right triangle?

Jlenok [28]

Answer:

{6, 8, 10} is a set which represents the side length of a right triangle.

Step-by-step explanation:

In a right triangle:

$(Base)^{2} + (Perpendicular)^{2} = (Hypotenuse)^{2}$

Now, in the given triplets:

(a) {4, 8, 12}

Here, $(4)^{2} + (8)^{2} = 16 + 64 = 80\\\implies H = \sqrt{80} = 8.94$

So, third side of the triangle 8.94 ≠ 12

Hence, {4, 8, 12} is NOT a triplet.

(b) {6, 8, 10}

Here, $(6)^{2} + (8)^{2} = 36 + 64 = 100\\\implies H = \sqrt{100} = 10$

So, third side of the triangle 10

Hence, {6, 8, 10} is a triplet.

Here, $(6)^{2} + (8)^{2} = 36 + 64 = 100\\\implies H = \sqrt{100} = 10$

So, third side of the triangle 10 ≠ 15

Hence, {6, 8, 15} is NOT a triplet.

(d) {5, 7, 13}

Here, $(5)^{2} + (7)^{2} = 25 + 49 = 74\\\implies H = \sqrt{74} = 8.60$

So, third side of the triangle 8.60 ≠ 13

Hence, {5, 7, 13} is NOT a triplet.

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3 years ago

Why do whole numbers raised to an exponent get greater well fractions raised to an exponent get smaller

adelina 88 [10]

Because when you raise an exponent on a fraction, it is basically dividing into itself. Imagine 1/3 multiplied my 3 (Ans is 1), the fractions aren't whole numbers so when they multiply into another number, it uses division

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3 years ago

Admission to the amusement park was $29.50 per person. if x represents the total cost for 3 tickets which equation is true?

Vesna [10]

Answer:

Cost per person × number of persons = total cost

29.50 × 3 = x

6 0

3 years ago

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Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$

$Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------$

$Answer:\\\large\boxed{-1+i}\\----------------------$

$Hope~This~Helped!~Good~Luck!$

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3 years ago

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