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3 years ago

Given side lengths of 40 cm, 70 cm, and 80 cm, how many triangles can be constructed?

Mathematics

2 answers:

Salsk061 [2.6K]3 years ago

7 0

One with 40cm being the side, 70 diagonal and 80 as the base so it is a right angle triangle

JulijaS [17]3 years ago

3 0

Answer:

The Answer is One :)

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What is the simplified expression for 44.4°? 49 40 41 42 43

SpyIntel [72]

Answer:

$4^2$

Step-by-step explanation:

$\dfrac{4^4 \cdot 4^3}{4^5}\\\\\\=\dfrac{4^{4+3}}{4^5}\\\\\\=\dfrac{4^{7}}{4^5}\\\\\\=4^{7-5}\\\\\\=4^2$

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2 years ago

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How do you solve 2 (+-1) =24 solve 2(+-1) =24

ExtremeBDS [4]

Answer:

2 + (-1) = 24

2-1 = 24

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Step-by-step explanation:

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3 years ago

What is the equation in point slope form for a linear function with a slope of m= -4 that passes through the point (2, -7)?

miskamm [114]

Answer:

-7-y1=-4(2-x1)

Step-by-step explanation:

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3 years ago

It is given that p varies inversely as q. If p = 12 and q = 45 find p, if q is 135.

alexgriva [62]

Answer:

$p =4$

Step-by-step explanation:

Given

Variation: Inversely

$p = 12$ when $q = 45$

Required

Determine p when $q = 135$

First, we need to determine the relationship between p and q

Since, it is an inverse variation.

The relationship is:

$p\ \alpha\ \frac{1}{q}$

Convert to an equation

$p= k *\frac{1}{q}$

$p= \frac{k}{q}$

Where k = constant of variation

Make k the subject

$k = p * q$

When $p = 12$ and $q = 45$

$k = 12 * 45$

$k = 540$

To solve for p when q = 135

Substitute 135 for q and 540 for k in $p= \frac{k}{q}$

$p = \frac{540}{135}$

$p =4$

6 0

2 years ago

What are the solutions of the equation (x + 2)2 – 2(x + 2) – 15 = 0? Use u substitution to solve.

Scrat [10]

Answer:

Step-by-step explanation:

hello :

(x + 2)2 – 2(x + 2) – 15 = 0

let : t= x+2

t²-2t-15 =0

(t-5)(t+3)=0

t-5=0 or t+3=0

t=5 or t=-3

but : t =x+2

if : t=5 x+2 =5 so : x=3

if : t= -3 x+2 = -3 so : x= -5

conclusion two solutions : 3 , -5

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3 years ago