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3 years ago

Help me with this question loviesss

Mathematics

1 answer:

alekssr [168]3 years ago

5 0

Answer: y = 3x - 1

Step-by-step explanation:

the line crosses the y axis at -1 and the x axis less than 1 .

y = 3x - 1

y = 0 - 1

y intercept = -1

0 = 3x - 1

3x = 1

x intercept = 1/3

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A right triangle has a leg of length 4.6 inches and a hypotenuse of length 5.4 inches. Find the approximate length of the other

SpyIntel [72]

The answer is 2.8in

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4 years ago

Madison and Tanisha went out to lunch. The bill came to $18.94. Madison knows that her lunch cost 9.72. How much did Tanisha's l

Finger [1]

Answer:

Tanisha's lunch cost $9.22.

Step-by-step explanation:

Given that,

Total bill = $18.94

Cost for Madison cost = $9.72

Total cost = Tanisha's lunch + Madison's lunch

18.94 = Tanisha's lunch + 9.72

Tanisha's lunch = 18.94 - 9.72

Tanisha's lunch = 9.22

So, Tanisha's lunch cost $9.22.

4 0

3 years ago

Whats -8,24,0,9,-13,17 in order from greatest to least

Lady bird [3.3K]

Answer:

-13, -8, 0, 9, 17, 24

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

1/7 times 4/5 I am not very good at fractions and I need help

frez [133]

1/7 * 4/5 = 4/35

Just multiply the numerators and the denominators as you would in an equation/expression.

1*4 = 4
7*5 = 35

4/35

8 0

3 years ago

Read 2 more answers

Calcula el cuadrado de cada binomio: A. (9 + 4m)^2

Olin [163]

Answer:

Part A) $(9+4m)^{2}=81+72m+16m^{2}$

Part B) $(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

Part C) $(2x-3z)^{2}=4x^{2}-12xz+9z^{2}$

Part D) $(4m^{5}+5n^{3})^{2}=16m^{10}+40m^{5}n^{3}+25n^{6}$

Part E) $(\frac{3}{6}w-\frac{1}{2}y)^{2}=\frac{9}{36}w^{2}-\frac{3}{6}wy+\frac{1}{4}y^{2}$

Part F) $(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

Step-by-step explanation:

The question in English is

Calculate the square of each binomial

we know that

The square of a binomial is always a trinomial

$(a+b)^{2}=a^{2}+2ab+b^{2}$

and

$(a-b)^{2}=a^{2}-2ab+b^{2}$

Part A) we have

$(9+4m)^{2}$

Applying the formula

$(9+4m)^{2}=(9)^{2}+2(9)(4m)+(4m)^{2}$

$(9+4m)^{2}=81+72m+16m^{2}$

Part B) we have

$(x^{10}-5y^{2})^{2}$

Applying the formula

$(x^{10}-5y^{2})^{2}=(x^{10})^{2}-2(x^{10})(5y^{2})+(5y^{2})^{2}$

$(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

Part C) we have

$(2x-3z)^{2}$

Applying the formula

$(2x-3z)^{2}=(2x)^{2}-2(2x)(3z)+(3z)^{2}$

$(2x-3z)^{2}=4x^{2}-12xz+9z^{2}$

Part D) we have

$(4m^{5}+5n^{3})^{2}$

Applying the formula

$(4m^{5}+5n^{3})^{2}=(4m^{5})^{2}+2(4m^{5})(5n^{3})+(5n^{3})^{2}$

$(4m^{5}+5n^{3})^{2}=16m^{10}+40m^{5}n^{3}+25n^{6}$

Part E) we have

$(\frac{3}{6}w-\frac{1}{2}y)^{2}$

Applying the formula

$(\frac{3}{6}w-\frac{1}{2}y)^{2}=(\frac{3}{6}w)^{2}-2(\frac{3}{6}w)(\frac{1}{2}y)+(\frac{1}{2}y)^{2}$

$(\frac{3}{6}w-\frac{1}{2}y)^{2}=\frac{9}{36}w^{2}-\frac{3}{6}wy+\frac{1}{4}y^{2}$

Part F) we have

$(x^{10}-5y^{2})^{2}$

Applying the formula

$(x^{10}-5y^{2})^{2}=(x^{10})^{2}-2(x^{10})(5y^{2})+(5y^{2})^{2}$

$(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

Note the problem F is the same problem B

5 0

3 years ago